1. ap chemistry chapter 7. buffers, titrations, and solubility
buffers and solubility handout (includes homework, schedule, outline, experiments, and problem set) lesson 1: introduction to buffers lesson 2: buffers and titration     solved problem lesson 3: pH and pKa lesson 4: titration and indicators lesson 5: introduction to solubility lesson 6 solubility equilibrium: Ksp lesson 7: pH and solubility buffers and solubility review     buffers AP released items with solutions     solubility AP released items with solutions ​buffers and solubility practice test

3. These dates are for 2016-2017- they will be updated as this unit approaches.

4. and buffers A weak acid or a weak base and 1: example: acetic acid
A buffer solution is a solution of: two things: 1:
A weak acid or a weak base
example:
acetic acid
CH3COOH
and
The common ion effect is
the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. 2:
The salt of the weak acid, which is a weak base
example:
sodium acetate
CH3COONa
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
why is this mixture special? Consider the reaction of each with water
consider the effect of adding adding acid to this mixture, noting that it has a lot of CH3CO2- in it
a weak acid
This is the big idea of a buffer. If you add the salt that an acid or base forms it will help prevent that acid from ionizing- ;e chateliers principle
CH3COOH (aq) + H2O D H3O+ (aq) + CH3COO- (aq)
and
It keeps the acid or base from ionizing, rendering it relatively harmless.
CH3COO- (aq) + H2O D CH3CO2H (aq) + OH-
a weak base
since the solution contains both a weak acid and a weak base, it will neutralize acids and bases
If we add a acid
common ion
H+ (aq) + CH3COO- (aq) D CH3COOH (aq)
H+ is neutralized by the weak base
…and if we add base:
OH- (aq) + CH3COOH (aq) D CH3COO- (aq) + H2O (l)
OH- neutralized by the weak acid
hopefully students will notice the bottom arguments are silly on the face of it because both reactions would happen if no sodium acetate were added, which seems to mean that acetic acid will not change it’s pH if you add acid or base!
The difference is that in a buffered solution we have a large excess of acetate (since we added sodium acetate) and acetic acid (since the equilibrium shifts to the left we will have more acetic acid present than in an unbuffered solution). So…it really does make sense.
Another way of looking at it. In the final reaction they have OH- reacting with acetic acid. Wait a minute….doesnt OH- react with H+ and that’s why adding base increases pH?? That’s exactly right and that’s why buffer systems don’t work with strong acids or bases. But we have a weak acid which means tons of Acetic acid and not much H+, so the hydroxide will tend to react with acetic acid and get absorbed into the collective…
therefore:
buffer solutions resist a change in pH
Adding product prevents the forward reaction from occuring
Add strong acid
Add strong base

5. Yes….pyridine (weak base) and its salt.
Is it a buffer solution?
They need a clear list of stong acids, strong bases, weak acids, and weak bases.
(a) KH2PO4/H3PO4
(b) NaClO4/HClO4
(c) C5H5N/C5H5NHCl
HCl/KCl
HCl/sodium acetate
Yes….phosphoric acid (weak acid) and its salt (potssium dihydrogen phosphate)
No….perchloric acid is strong…perchorate (HClO4- is so weak it won’t react with H+)
Yes….pyridine (weak base) and its salt.
no because
HCl is a strong acid (complete dissociation, no equilibrium)
Cl- is a neutral ion (will not react with H+)
yes because it reacts to make the weak acid needed
H+ + CH3CO2-  CH3CO2H

6. x = [H3O+] =2.1 x 10−5M; pH = −log (2.1 x 10−5) = 4.68
What is the pH of a 0.700M acetic acid/0.600M sodium acetate buffer system? The Ka of acetic acid is 1.8 x Compare this to the pH without buffer (acetate) present
a “pH of the buffer system” problem
fast:
ice table method
CH3CO2H(aq) D
H+(aq) +
CH3CO2-(aq)
CH3CO2H(aq) D
H+(aq) +
CH3CO2-(aq)
initial
0.7
0.6
change -x +x
equilibrium
0.7-x =0.7 x
0.6+x = 0.6
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.600] [0.700]
x = [H3O+] =2.1 x 10−5M; pH = −log (2.1 x 10−5) = 4.68
(as usual we can simplify this since 0.7 and 0.6 are so much larger than 10-5)
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.600] [0.700]
x = [H3O+] =2.1 x 10−5M; pH = −log (2.1 x 10−5) = 4.68
Compare to pH without buffer: 1.8 x 10-5 = x2/0.7M;
x = [H+] = M;
pH = -log = 2.45…makes sense

7. Ka = 1.4 x 10-4 = H+ [L−] [LH] = x [0.025] [0.025] ;
What is the pH of the solution that results from adding 25.0 mL of M NaOH to mL of 0.100M lactic acid? The Ka of lactic acid is 1.4 x 10-4.
a “reaction that makes a buffer” problem; note excess lactic acid
This reaction will create a buffer system of lactic acid (HL) and sodium lactate (NaL). Find the concentration of each then solve like the last one.
HL NaOH  H2O NaL
0.025L x mol L
= mol
0.025L x mol L
= mol
mol
the reaction of 25 mmol lactic acid with 12.5 mmol NaOH will make 12.5 mmol of sodium lactate, with mmol of lactic acid remaining.
since there are 50 mL of solution, the concentration of each is mol/.05 L = 0.025M
now we have our buffer system- it is 0.025M lactic acid/0.025M sodium lactate. Lets find the pH
LH(aq) D
H+(aq) +
L-(aq)
initial
0.025
change -x +x
equilibrium
0.025-x =0.025 x
x = 0.025
faster: skip ice table)
Ka = 1.4 x 10-4 = H+ [L−] [LH] = x [0.025] [0.025] ;
x = [H3O+] =1.4 x 10−4M;
pH = −log (1.4 x 10−4) = 3.85

8. Titration What happens in a titration: Equivalence point: Indicator:
The concentration of an unknown acid or base is determined by neutralizing it with a measured amount of acid or base (usually base).
If they are both strong acids and bases this can be solved using C1V1 = C2V2. For weak acids (or bases) the Ka of the acid (or occasionally the Kb of the base) comes into play.
Equivalence point:
the point at which the reaction is complete.
Also the midpoint in the vertical portion of the pH curve
Indicator:
substance that changes color at (or near) themequivalence point
Procedure:
Slowly add base to unknown acid until the indicator changes color (pink)
monitor pH
alternative procedure:

9. titrations: four scenarios
pH at equivalence point
pH after equivalence point
initial pH and how to measure
pH as titant is added before equivalence point…is buffer created?
plot and example
low: = pH [H+] = - log [HX]
7: forms a neutral salt
high
strong acid strong base added
H+ consumed…slow increase
1. H+ is slowly consumed by NaOH
2. All H+ is consumed
NaOH (aq) + HCl (aq)  H2O (l) + NaCl (aq)
3. OH- is added but doesn’t react, increasing pH
OH- (aq) + H+ (aq)  H2O (l)
not as low pH can be calc. from ka and [H+]
>7: forms a basic salt
high
H+ consumed…buffer is created. Slow inc.
weak acid strong base added
CH3COOH (aq) + NaOH (aq)  CH3COONa (aq) + H2O (l)
the half equivalence point is also very significant here (see next slide)
CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + H2O (l)
weak base strong acid added
<7: forms a acidic salt
high
base consumed..buffer is created
low
HCl (aq) + NH3 (aq)  NH4Cl (aq)
H+ (aq) + NH3 (aq)  NH4+ (aq)
generally equivalent to 2 or 3 since one will be stronger than the other.
Weak acid weak base

10. the halfway point for weak acid/strong base titrations
(also known as the half-equivalence point)
reaction is halfway done- note the volumes.
ex: acetic acid and NaOH (ka = 1.8 x 10-5, so pKa = 4.74)
at the halfway point pH = pKa
and [H+] = Ka
what is the [H3O+] at the halfway point for the titration of acetic acid with NaOH?
1.8 x 10-5 M
what is the pH at the halfway point for the titration of acetic acid with NaOH?
4.74

11. pick a partner
draw in each for future years…pretty easy
draw a titration curve from the choices below. Include
1. labeled x and Y axes with numbers
2. the endpoint
3. the halfway point
orally defend your starting and finishing pH’s and the pH at your endpoint pH
“our starting pH is ____ because___”
“our finishing pH is ____ because___”
“our endpoint pH is ____ because___”
also note any other points such as the halfway point.
1. weak acid titrated with a strong base
2. strong acid titrated with a strong base
3. strong acid titrated with a weak base
4. strong base titrated with a strong acid

12. The titration curve of a strong acid with a strong base.
Acid-Base Indicators
HIn (aq) H+ (aq) + In- (aq)
 10
[HIn]
[In-]
Color of acid (HIn) predominates
 10
[HIn]
[In-]
Color of conjugate base (In-) predominates
The titration curve of a strong acid with a strong base.

13. equivalence point versus endpoint
equivalence point is determined theoretically
endpoint is measured experimentally
sample question
A scientist for NASA is asked to find the pH of a strange pool of liquid returned
from Io, one of saturns moons. Using a pH meter he measures the pH to be 0.51.
He confirms this by titrating it with NaOH, thus he is titrating a _______ acid with a ___________ base. He calculates that [H3O+] is ______M, and he assumes that it will take mL of NaOH of the same molarity, to neutralize mL of the unknown liquid, that is the calculated. When he performs the titration using phenolphthalein as an indicator he finds that it takes 20.3 mL of the base to neutralize it. Explain.
strong
strong
0.309
He calculated the equivalence point (20.0 mL of 0.309M NaOH), and measured the endpoint (20.3 mL of 0.309M). Since phenolphthalein undergoes a pH change at this is not surprising. The NASA scientist should have used a pH meter since a titration of a strong base with a strong acid will have a pH of 7.00 at it’s endpoint

14. Solubility Equilibria
similarly
MgF2 (s)D Mg2+ (aq) + 2F- (aq)
consider dissolving silver chloride in water:
AgCl (s) D Ag+ (aq) + Cl- (aq)
no units
Ksp = [Mg2+][F-]2 = 7.4 x 10-11
Ksp = [Ag+][Cl-]
why no denominator I Ksp?
solid. No equilibrium
Ksp is the solubility product constant
thus far I see solubility product constants as a very bad way of measuring solubility.
where [Ag+] and [Cl-] are each the molar concentrations of Ag+ and Cl- in water at 25 OC.
utility of Ksp:
we can use it to predict precipitates. Recall that Q is a measured equilibrium compared to the known equilibrium and that
if Q< Keq the reaction shifts >
if Q = Keq the reaction is at equilibrium
and if Q > Keq the reaction shifts <
since the solubility of silver chloride in water is 1.26 x 10-5 mol/L, Ksp for AgCl = ______.
1.6 x 10-10
why not skip all this mumbo jumbo and look up solubilities in g/L??
similarly,
Q < Ksp
Unsaturated solution
No precipitate
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
Precipitate will form
adding either ion will affect solubility (the common ion effect)
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2

15. sample problems: 17,18 (go to- they are solved)

16. pH and Solubility
The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
add
remove
Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
What can Ksp of Mg(OH)2 tell us?
At pH less than 10.45
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Lower [OH-]
Ksp = (s)(2s)2 = 4s3
OH- (aq) + H+ (aq) H2O (l)
4s3 = 1.2 x 10-11
Increase solubility of Mg(OH)2
s = 1.4 x 10-4 M
At pH greater than 10.45
[OH-] = 2s = 2.8 x 10-4 M
Raise [OH-]
pOH = pH = 10.45
Decrease solubility of Mg(OH)2
Why the difference?
substance
molar solubility
factor to get Ksp
Ksp calculated
Ksp published
AgBr
7.35 x 10-7 x2
5.40 x 10-13
7.7 x 10-13
They went a bit deeper. The Ksp values have been adjusted to accurately measure the concentrations of ions in solution, but more may dissolve undissociated, and other factors may come into play as well.

17. KEY still needs lots of work
buffers
chapter
problems
KEY still needs lots of work

18. 1. What is the pH of a 0. 700M lactic acid/0
1. What is the pH of a 0.700M lactic acid/0.600M sodium lactate buffer system? The Ka of lactic acid is 1.4 x Compare this to the pH without buffer (lactate) present
2. What is the pH of the solution that results from adding 25.0 mL of M NaOH to mL of 0.100M acetic acid? The Ka of acetic acid is 1.8 x 10-5.
Note: acids were switched! Method is good but values are wrong…
a “reaction that makes a buffer” problem; note excess acetic acid
lactic acid is HC3H5O3, so the equilibrium reaction is:
HC3H5O3 (aq) D H+(aq) = C3H5O3-
with lactate we have a buffer system so H-H equation is easiest
pH = pKa + log [A−]l [HA] = -log (1.4 x 10-4) + log [0.600]l [0.700] = 3.79
without buffer we have to solve conventionally:
1.4 x 10-4 = x2/0.7M;
x = [H+] = M;
pH = -log = 2.00…makes sense
This reaction will create a buffer system of acetic acid and sodium acetate. Find the concentration of each then solve like the last one.
fast:
HC2H3O NaOH D H2O NaC2H3O2
0.025L x mol L
= mol
0.025L x mol L
= mol
CH3CO2H(aq) D
H+(aq) +
CH3CO2-(aq)
mol
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.600] [0.700]
the reaction of mol of acetic acid with mol NaOH will make mmol of sodium acetate, with mol of acetic acid remaining. (You may notice the pH will equal the pKa)
x = [H3O+] =2.1 x 10−5M; pH = −log (2.1 x 10−5) = 4.68
ice table method
since there are 50 mL of solution, the concentration of each is mol/.05 L = 0.025M
CH3CO2H(aq) D
H+(aq) +
CH3CO2-(aq)
initial
0.7
0.6
change -x +x
equilibrium
0.7-x =0.7 x
0.6+x = 0.6
pH = pKa + log [A−]l [HA] = -log (1.8 x 10-5) + log [0.025]l [0.025] = 4.74
(as usual we can simplify this since 0.7 and 0.6 are so much larger than 10-5)
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.600] [0.700]
x = [H3O+] =2.1 x 10−5M; pH = −log (2.1 x 10−5) = 4.68
Compare to pH without buffer: 1.8 x 10-5 = x2/0.7M;
x = [H+] = M;
pH = -log = 2.45…makes sense
now we have our buffer system- it is 0.025M lactic acid/0.025M sodium lactate. Lets find the pH
LH(aq) D
H+(aq) +
L-(aq)
initial
0.025
change -x +x
equilibrium
0.025-x =0.025 x
x = 0.025 18
Ka = 1.4 x 10-4 = H+ [L−] [LH] = x [0.025] [0.025] ;
x = [H3O+] =1.4 x 10−4M;
pH = −log (1.4 x 10−4) = 3.85

19. 3. Indicate and explain the direction of pH change when
solid disodium oxalate (Na2C2O4) is added to 50 mL of 0.015M oxalic acid (H2C2O4).
4. What is the pH after 1.56 g of sodium acetate is added to 100 mL of 0.15M acetic acid? Note that the Ka of acetic acid is 1.8 x 10-5.
[CH3CO2-] = 1.56g 0.1L x mol 82.0 g = 0.190M
pH will increase since oxalate is a weak base
CH3CO2H(aq) D
H+(aq) +
CH3CO2-(aq)
initial
0.15
0.190
change -x +x
equilibrium
0.15 –x = 0.15 x
0.190-x = 0.190
b. solid ammonium chloride is added to 75 mL of 0.016M HCl
pH will increase slightly since NH4+ is a weak acid
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.190] [0.15]
x = 1.4 x 10−5M; pH = −log (1.4 x 10−5M) = 4.85
c. 20 g of NaCl is added to 1.0L of 0.10M sodium acetate
pH will not change because NaCl is a neutral salt
5. What is the pH of the solution after 25.O mL of 0.12M HCl is combined with 25 mL of 0.43M NH3? Note that the Ka of NH4+ is 5.6 x
NH3 + HCl  NH4Cl
NH4+(aq) D
H+(aq) +
NH3(aq)
initial
0.060
0.20M
change -x +x
equilibrium
–x = 0.060 x
0.20-x = 0.20
NH3: (0.025L)(0.43mol/L) = mol
HCl: (0.025L)(0.12mol/L) = mol
this will make mol NH4+ with mol NH3 unreacted
Ka 5.6 x = H+ [NH3] [NH4+] = x [0.20] [0.06]
x = 2.0 x 10−10 M; pH = −log (2.0 x 10−10 M) = 9.7
molarities (note the volume of the solution is 50 mL = 0.050L)
NH3: mol/0.050L = 0.20M
NH4+: mol/0.050L = M

20. = 49.6 g sodium lactate per liter of buffer solution
8. A 1.000L buffer solution is composed of 1.860g of KH2PO4 and 5.977g of Na2HPO4. Note that the Ka of KH2PO4 is 6.2 x 10-8.
What is the pH of the buffer solution?
Suggest a method to lower the pH of this buffer solution by 0.8 units.
6. What is the pH of the buffer solution. that contains 2.2 g of NH4Cl in 250 mL of 0.12M NH3? Is the final pH lower or higher than the pH of the 0.12M ammonia solution? Note that the Ka of NH4+ is 5.6 x
check with answer key and see if H-H is an easier solution; consider showing both if it is
this is the long way for 6
NH4+(aq) D
H+(aq) +
NH3(aq)
a. H2PO4- D HPO42- + H+; Ka = 6.2 x 10-8 = HPO42− [H+] [H2PO4−]
solve using H-H equation. We need [HA] and [A-]:
[HA] = [NH4] = [NH4Cl]
= 2.2 g NH4Cl x mol NH4CL 53.5 g NH4Cl x 1 [025 L solution = M
[H2PO4-] = [KH2PO4]
= g KH2PO4 L x mol KH2PO g KH2PO4 = M
NH4+(aq) D
H+(aq) +
NH3(aq)
initial
0.160
0.12M
change -x +x
equilibrium
–x = 0.160 x
0.120-x = 0.120
[A-] = [NH3] = .12M
[HPO42-] = [Na2HPO4]
= g Na2HPO4 L x mol Na2HPO g Na2HPO4 = M
pH = pKa + log [A−]l [HA] = -log (5.6 x 10-10) + log [.12]l [.164] = 9.11
EASIER to do HH
Ka 5.6 x = H+ [NH3] [NH4+] = x [0.120] [0.160]
x = x 10−10 M; pH = −log (7.7.0 x 10−10 M) = 9.11
the addition of the weak acid NH4+ will lower the pH.
7. How much sodium lactate must be added to 1.00L of 0.10M lactic acid to give the solution of pH 4.50? Note that the Ka of lactic acid is 1.4 x 10-4.
5.6 x = [x] [ ] ;x = H+ = x 10−8;
pH = -log x 10-8 = 7.81
Hlactate D = H+ + lactate-
b. will lower the pH by adding KH2PO4.
[H+] = 10-pH = = x 10-8
4.5 = -log(1.4 x 10-4) + log [lactate−] [0.1]
[NH4Cl] = [NH4+] = 12.2g 0.25L x mol 53.5 g =0.160M
6.2 x 10-8 = [4.898 x 10−8] [x]
0.646 = log [lactate−] [0.1]
x = [H2PO4-] = [KH2PO4] = M
4.426 = [lactate−] [0.1]
molKH2PO4 l x g KH2PO4 mol KH2PO4
= 4.29 g KH2PO4
4.29 g KH2PO4 – g KH2PO4 =
add 2.94 g KH2PO4
𝑙𝑎𝑐𝑡𝑎𝑡𝑒− = mol lactate 1.0 liter 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 x mol Nalactate mol lactate− x g Nalactate mol Nalactate
= 49.6 g sodium lactate per liter of buffer solution
[H+] = 10-pH = = 3.2 x 10-5M
Ka = 1.4 x 10-4 = H+ [lactate−] [Hlactate] = 3.2 x 10−5 M [x] [0.10]
x =[lactate−] = 0.057M

21. these are problem 20 and 22 from kotz ch 18
g of NaOH is dissolved in 2.00L of a buffer solutions where both [H2PO4-] and [HPO42-] are 0.132M. What is the pH before adding NaOH. What is the pH after adding NaOH?
10. What will the pH change to when 20 mL of 0.100M NaOH is added to 80 mL of a buffer solution consisting of 0.169M NH3 and 0.183M NH4Cl?
these are problem 20 and 22 from kotz ch 18

22. 13. Calculate the pH in the titration of 25. 0 mL of 0
13. Calculate the pH in the titration of 25.0 mL of M acetic acid by sodium hydroxide after the addition of 25.0 mL of M NaOH. The Kb of CH3CO2- is 5.6 x 10-10
11. What is the pH of a solution after 10.0 mL of 0.100M NaOH has been added to 50.0 mL of 0.100M HCl? Solve by filling in the blanks.
Fast method: reaction will produce mol of CH3CO2Na in 50 mL of solution = 0.05M
CH3CO2- + H2O D CH3CO2H + OH-
Kb = 5.6 x = CH3CO2H [OH−] [CH3CO2−] = x −x = x
x =[OH−] 5.3 x10−6; ; pOH = −log5.3 x10−6= 5.28;
pH = 14 – pOH = 8.72
type of problem:
partial titration of strong acid with strong base
H+(aq) OH- (aq)  H2O (l)
mol
0.005
______
0.001
______
when the reaction is over there will be __________ mol of H+, __________mol of OH (and __________ mol of water )
.00400
Slow method:
acetate is a weak base in equilibrium with water
since the final volume of the solution after the reaction is over is _______ L, the [H+] is ____________M
which means the pH is ___________
.060
.0667
1.176
CH3CO2H(aq) +
OH- 
CH3CO2-(aq)
initial mol
.00250
change
-.0025
.001
final
12. Calculate the pH in the titration of 25.0 mL of M acetic acid by sodium hydroxide after the addition of 10.0 mL of M NaOH. Note that the Ka of acetic acid is 1.8 x 10-5.
CH3CO2-(aq) + H2O D
CH3CO2H +
OH-
[initial]
.00250mol/.05L = .0500M
[change]
- x
+ x
[final] x
CH3CO2H(aq) +
OH- 
CH3CO2-(aq)
initial mol
.00250
change
-.001
.001
final
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [ ] [ ]
x =[H+] = 2.70 x 10−5; pH = -log (2.7 x 10-5) = 4.57
Kb = 5.6 x = CH3CO2H [OH−] [CH3CO2−] = x −x = x
x =[OH−] 5.3 x10−6; ; pOH = −log 5.3 x 10−6 = 5.28;
pH = 14 – pOH = 8.72

23. we can ignore the acetate since the hydroxide overwhelms it.
14. Calculate the pH in the titration of 25.0 mL of M acetic acid by sodium hydroxide after the addition of 35.0 mL of M NaOH.
CH3CO2H(aq) +
OH- 
CH3CO2-(aq)
initial mol
.00250
change
-.0025
final
we can ignore the acetate since the hydroxide overwhelms it.
note that the volume of the solution is 60 mL
[OH-] = mol 0.060L = M
pOH = = 1.78
pH = 14 – = 12.22

24. A solution of 25.0 mL of 0.10M NH3 is titrated with 0.10M HCl.
what is the pH of the NH3 solution before the titration begins? The Kb of NH3 is 1.8 x 10-5.
what is the pH at the equivalence point? The Ka of NH4+ is 5.6 x
What is the pH at the halfway point (also known as the half-equivalence point) of the titration?
Suggest a suitable indicator for this titration
Calculate the pH of the solution after adding 5.00, 15.00, 20.00, 22.0, and 30 mL of the acid. Plot the titration curve on the graph below.
a. NH3 is a weak base:
(b, continued)
Ka = 5.6 x = = [H+][[NH3] [NH4+] = x
x = [H+] = 5.3 x 10-4M
pH = -log 5.3 x 10-4 = 5.28
c. at the half equivalence point [NH3] = [NH4+] and pH = pKa = -log (5.6 x 10-10) = 9.25
d. Methyl red works well at a pH of 5.28.
e. (work not shown)
5.00 ml: pH = 9.85, ml pH = 9.08, 20.0 mL pH = 8.65, 22.0 mL pH = 8.39, 30.0 mL pH = 2.04.
NH3(aq) + H2O(l) D
NH4+(aq) +
OH-(aq)
initial M
0.10
change -x +x
equilibrium
0..10 –x = 0.10 x
Kb = 1.8 x 10-5 = = [NH4+][[OH−] [NH3] = x2 0.10
x = [OH-] = M
pOH = -log = 2.87
pH = = 11.13
b. At the equivalence point 25 mL of 0.10M NH3 has reacted with 25 mL of 0.10M HCl L x mol L = mol of NH4+ has formed and there is now 50 mL of solution: mol L =0.050M
NH4+(aq) D
H+(aq) +
NH3(aq)
initial M
0.050
change -x +x
equilibrium
– x = 0.050 x

25. pH will be dominated by the excess strong acid
Question 15 part E
A solution of 25.0 mL of 0.10M NH3 is titrated with 0.10M HCl.
e. Calculate the pH of the solution after adding 5.00, 15.00, 20.00, 22.0, and 30 mL of the 0.1M HCl.
NH3
starting
HCl
NH4+
produced
NH3 remaining
pH = pKa + log A-/HA M L
mol
pKa = -log 5.6 x = 9.25
0.1
.005
.0005
.025
.0025
.030
.0167
.002
.0667
= log .0667/.0167
= 9.85
.015
.0015
0025
.050
.0010
.033
= log ..033/.050
= 9.07
.020
.0020
= log .0167/.0667
= 8.64
.022
.0022
.0733
.0003
.01
= log .010/.0733
= 8.38
.0030
pH will be dominated by the excess strong acid
[H+] = mol/.055L
= M
-log = 2.04

27. 16. Which indicator or indicators listed in Table 17
16. Which indicator or indicators listed in Table 17.1 would you use for the acid-base titrations shown?
(a) Near the equivalence point, the pH of the solution changes abruptly from 4 to 10. Therefore, all the indicators except thymol blue, bromophenol blue, and methyl orange are suitable for use in the titration.
(b) Here the steep portion covers the pH range between 7 and 10; therefore, the suitable indicators are cresol red and phenolphthalein.
(c) Here the steep portion of the pH curve covers the pH range between 3 and 7; therefore, the suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue.

28. All questions changed…these answers may be off, let me know
17. Calculate the solubility of chalk (calcium carbonate) in moles per liter and grams per liter. Its Ksp is 8.7 x The molecular weight of CaCO3 is 100. g/mol.
20. The solubility of calcium sulfate (CaSO4) is found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate. (CaSO4 has a molar mass of g/mol)
CHANGE TO CaCO3
BaSO4(s) D Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][SO42-] = 1.1 x 10-10
= (s)(s) where s = the solubility of BaSO4
s = x 10−10 = 1.0 x 10-5M
1.0 x 10−5 mol L x 233g mol = g/L
CaSO4(s) D Ca2+(aq) + SO42-(aq)
0.67 g L x mol g = M = s
Ksp = [Ca2+] ][SO42-] = s2 = (4.9 x 10-3 )(4.9 x 10-3 ) = 2.4 x 10-5
18. Calculate the Ksp for CaF2 if the calcium ion concentration has been found to be 1.3 x 10-4M.
21. A scientist places 5.00 micrograms of copper(II)hydroxide in one liter of water. After 24 hours, he filters the solution and finds that 3.25 milligrams of the substance remains undissolved. What is the percent error of his measurement if the Ksp of Cu(OH)2 is 2.2 x 10-20?
Fast: 4(1.3 x 10-4)3 = 8.8 x 10-12
slower
CaF2(s) D Ca2+(aq) + 2F-(aq)
Ksp = [Ca2+][F-]2 = (s)(2s)2 = 4s3
Ksp = (1.3 x 10-4)(2.6 x 10-4)2 = 8.8 x 10-12
s = x 10− =
19. Calculate the solubility of Mg(OH)2 in moles per liter and grams per liter. Its Ksp is 1.2 x and molecular weight is 58.3 g/mol.
Mg(OH)2(s) D Mg2+(aq) + 2OH-(aq)
Ksp = 1.2 x = [Mg2+][OH-]2 = (s)(2s)2 = 4s3
s = x 10− = 1.4 x 10-4M
1.4 x 10−4 mol L x g mol = 8.4 x 10-3 g/L
22. Which substance has a higher solubility in water in g/L, barium fluoride (Ksp = 1.7 x 10-6) or barium sulfate (Ksp = 1.1 x 10-10)? Defend your answer.
All questions changed…these answers may be off, let me know

29. Yes a precipitate will form.
23. Exactly mL of M BaCl2 are mixed with mL of M K2SO4. Will a precipitate form? The Ksp of BaSO4 is 1.1 x 10-10
24. Calculate the solubility of silver chloride (in g/L) in a
6.5 x 10-3 M silver nitrate solution. (Ksp of AgCl = 1.6 x )
BaSO4(s) D Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][SO42-] = 1.1 x 10-10
AgNO3(s) D Ag+(aq) + NO3-(aq)
6.5 x 10-3 M x 10-3 M
assume [Ag+] from AgCl dissolving is negligible
[Ba2+]: mol Ba2+ L x L = mol Ba2+
mol Ba L solution = [Ba2+] = M
Ksp AgCl = 1.6 x = [Ag+][Cl-] = (6.5 x 10-3 )s
s = 2.5 x 10-8 M
[SO42-]: mol SO42−+ L x L = mol SO42-
mol SO42− L solution = [Ba2+] = M
all chloride ions must come from AgCl so solubility of AgCl in this solution is 2.5 x 10-8 M
Q = [Ba2+][SO42-] = (1.0 x 10-3)(6.0 x 10-3)
= 6.0 x 10-6…too much! …
Yes a precipitate will form.
Or in more detail: Q > Ksp
The solution is supersaturated because the value of Q indicates that the concentrations of the ions are too large. Thus, some of the BaSO4 will precipitate out of solution until [Ba2+][SO42-] = [1.0 x 10-10]
P13. How many grams of AgCl can dissolve in 1L of a 0.55M NaCl solution?
Kotx 880
should point out that we cant simply compare the amount of barium sulfate produced to it’s known solubility because there is excess sulfate. That common ion will shift the solubility to the left.
key completed to here

30. P15. Solid AgCl has been placed in a beaker of water
P15. Solid AgCl has been placed in a beaker of water. After some time, the concentration of Ag+ and Cl- are each 1.2 x 10-5M. Has the system reached equilibrium? If not, will more AgCl dissolve?
Ksp AgCl = 1.8 x 10-10
Kotz 885
P16. The concentration of Ba2+ in a solution is 0.010M
What concentration of SO42- is required to just begin precipitating BaSO4?
When the concentration of sulfate ion in the solution reaches 0.015M, what concentration of barium ion will remain in the solution?
Ksp BaSO4 =1.1 x 10-10
Kotx 886

31. Ksp = 2.0 x 10-13

33. Using data from Table 17.2, calculate the solubility of CaF2 in g/L.
17.40
Using data from Table 17.2, calculate the molar solubility of calcium phosphate, which is a component of bones.
17.32
Why do we usually not quote the Ksp values for soluble ionic compounds?
17.32 Because they have very large Ksp values.
17.36
Silver chloride has a larger Ksp than silver carbonate (see Table 17.2). Does this mean that the former also has a larger molar solubility than the latter?
17.36 No. When comparing Ksp values to predict molar solubility, it is important to compare salts with similar subscripts in their formulas.
17.42
Using data from Table 17.2, calculate the solubility of CaF2 in g/L.
17.38
From the solubility data given, calculate the solubility products for these compounds:
SrF2, 7.3 × 10−2 g/L
Ag3PO4, 6.7 × 10−3 g/L

34. Ksp is the same as solubility.
17.48
The molar solubility of AgCl in 6.5 × 10−3 M AgNO3 is 2.5 × 10−8 M. In deriving Ksp from these data, which of these assumptions are reasonable?
Ksp is the same as solubility.
Ksp of AgCl is the same in 6.5 × 10−3 M AgNO3 as in pure water.
Solubility of AgCl is independent of the concentration of AgNO3.
[Ag+] in solution does not change significantly on the addition of AgCl to 6.5 × 10−3 MAgNO3.
[Ag+] in solution after the addition of AgCl to 6.5 × 10−3 M AgNO3 is the same as it would be in pure water.
17.44
The pH of a saturated solution of a metal hydroxide MOH is Calculate the Ksp for the compound.
17.44 First we can calculate the OH concentration from the pH.
pOH=  pH
pOH=  9.68 = 4.32
[OH] = 10pOH= 104.32= 4.8  105M
The equilibrium equation is
MOH(s)  M+(aq) + OH(aq)
From the balanced equation, we know that [M+] = [OH]
Ksp= [M+][OH] = (4.8  105)2=2.3  109
17.48 (b) and (d) are reasonable assumptions
17.46
A volume of 75 mL of 0.060 M NaF is mixed with 25 mL of 0.15 M Sr(NO3)2. Calculate the concentrations in the final solution of   and F−. (Ksp for SrF2 = 2.0 × 10−10.)
17.50
The solubility product of PbBr2 is 8.9 × 10−6. Determine the molar solubility (a) in pure water, (b) in 0.20M KBr solution, (c) in 0.20 M Pb(NO3)2 solution.

35. 17.50
The solubility product of PbBr2 is 8.9 × 10−6. Determine the molar solubility (a) in pure water, (b) in 0.20M KBr solution, (c) in 0.20 M Pb(NO3)2 solution. Explain

36. 17.52
Calculate the molar solubility of BaSO4 (ksp = 1.1 x 10-10)
in water and
(b) in a solution containing 1.0 M SO42- ions. Explain

37. 1972
a. Given a solution of ammonium chloride, what additional reagent or reagents are needed to prepare a buffer from the ammonium chloride solution
Since ammonium chloride (NH4Cl) is the salt of a weak base, the weak base NH3 is needed.
Similar explanation: the ammonium cation is a weak acid, so its conjugate base (NH3) is needed.
Explain how this buffer solution resists a change in pH when
b. Moderate amounts of strong acid are added
Moderate amounts of strong acid will react with the weak base NH3 to make the ammonium ion:
NH3 + H+ D NH4+
The concentration of hydrogen ion remains essentially the same resulting in only a very small change in pH
c. Moderate amounts of strong base are added
Moderate amounts of strong base will react with the ammonium cation
NH4+ + OH- D NH3 + H2O
The concentration of hydroxide ion remains essentially the same resulting in only a very small change in pH
d. A portion of the buffer solution is diluted with an equal volume of water.
By diluting with water the relative concentration of NH3/NH4+ does not change so there should be no change in pH

38. 1983
Specify the properties of a buffer solution. Describe the components and the composition of effective buffer solutions.
b. An employer is interviewing four applicants for a job as a laboratory technician and asks each how to prepare a buffer solution with a pH close to 9.
Archie A. says he would mix acetic acid and sodium acetate solutions.
Beula B. says she would mix NH4Cl and HCl solutions.
Carla C. says she would mix NH4Cl and NH3 solutions.
Dexter D. says he would mix NH3 and NaOH solutions.
Which of these applicants has given an appropriate procedure? Explain your answer, referring to your discussion in part (a). Explain what is wrong with the erroneous procedures. (No calculations are necessary, but the following acidity constants may be helpful: acetic acid, Ka = 1.8 × 10–5 NH4+, Ka = 5.6 × 10–10)
three points A buffer solution resists changes in pH upon the addition of an acid or base. Preparation:
• Mix a weak acid + a salt of a weak acid. • Or mix a weak base + a salt of a weak base. • Or mix a weak acid with about half as many moles of strong base. • Or mix a weak base with about half as many moles of strong acid. • Or mix a weak acid and a weak base.
five points Carla has the correct procedure. She has mixed a weak base, NH3, with the salt of a weak base, NH4Cl. Archie has buffer solution but it has a pH around 5. Beula does not have a buffer solution, since her solution consists of a strong acid and a salt of a weak base. Dexter does not have a buffer solution, since his solution consists of a weak base plus a strong base

39. 1988
Explain how this curve could be used to determine the molarity of the acid.
(b) Explain how this curve could be used to determine the dissociation constant Ka of the weak monoprotic acid.
(c) If you were to repeat the titration using a indicator in the acid to signal the endpoint, which of the following indicators should you select? Give the reason for your choice.
Methyl red Ka = 1 × 10–5 Cresol red Ka = 1 × 10–8 Alizarin yellow Ka = 1 × 10–11
(d) Sketch the titration curve that would result if the weak monoprotic acid were replaced by a strong monoprotic acid, such as HCl of the same molarity. Identify differences between this titration curve and the curve shown above.
Average score = 1.76
two points
The sharp vertical rise in the pH on the pH-volume curve appears at the equivalence point (about 23 mL). Because the acid is monoprotic, the number of moles of acid equals the number of moles of NaOH. That number is the product of the exact volume and the molarity of the NaOH. The molarity of the acid is the number of moles of the acid divided by L, the volume of the acid.
At the half-equivalence point (where the volume of the base added is exactly half its volume at the equivalence point), the concentration [HX] of the weak acid equals the concentration [X¯] of its anion. Thus, in the equilibrium expression, [H+] [X¯] / [HX] = Ka. Therefore, pH at the half-equivalence point equals pKa
Cresol red is the best indicator because its pKa (about 8) appears midway in the steep equivalence region. This insures that at the equivalence point the maximum color change for the minimal change in the volume of NaOH added is observed.

40. 1992
The equations and constants for the dissociation of three different acids are given below. HCO3– p H+ + CO32– Ka = 4.2 × 10–7 H2PO4– p H+ + HPO42– Ka = 6.2 × 10–8 HSO4– p H+ + SO42– Ka = 1.3 × 10–2
From the systems above, identify the conjugate pair that is best for preparing a buffer with a pH of 7.2. Explain your choice.
(b) Explain briefly how you would prepare the buffer solution described in (a) with the conjugate pair you have chosen.
(c) If the concentrations of both the acid and the conjugate base you have chosen were doubled, how would the pH be affected? Explain how the capacity of the buffer is affected by this change in concentrations of acid and base.
PH not changed. Capacity of buffer would because there are more moles of conjugate acid and conjugate base to react with added base on acid.
(d) Explain briefly how you could prepare the buffer solution in (a) if you had available the solid salt of the only one member of the conjugate pair and solution of a strong acid and a strong base.
Best conjugate pair: H2PO4- , HPO42-
When 7.2=PH=Pka for this pair when [HPO42-] = [H2PO4-]
Dissolve equal amounts of H2PO4- and HPO42- in water
Add strong base to salt of conjugate acid or add strong acid to salt of conjugate base. Add 1 mole conjugate acid to ½ mole strong base or one mole conjugate base to ½ mole strong acid or use PH meter to monitor addition of strong base to conjugate acid or strong acid to conjugate base.

41. 1998
An approximately 0.1-molar solution of NaOH is to be standardized by titration. Assume that the following materials are available.
Clean, dry 50 mL buret Analytical balance 250 mL Erlenmeyer flask Phenolphthalein indicator solution Wash bottle filled with distilled water Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard)
Briefly describe the steps you would take, using materials listed above, to standardize the NaOH solution. 4 essential steps.
(b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution.
(c) After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve.
(d) Describe how the value of the acid-dissociation constant, Ka, for the weak acid HX could be determined from the titration curve in part (c). (e) The graph below shows the results obtained by titrating a different weak acid, H2Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve.
Weight KHP
Fill buret with NaOH solution
Add indicator (Phenolphthalein)
Titrate to endpoint (color change)

42. 2001
Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.
The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of 2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet.
b. The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The combustion of g of the pure compound yields g of water and 3.72 L of dry carbon dioxide, measured at 750. mm Hg and 25°C. Calculate the mass, in g, of each element in the g sample.
c. A student dissolved g of pure acetylsalicylic acid in distilled water and titrated the resulting solution to the equivalence point using mL of M NaOH(aq). Assuming that acetylsalicylic acid has only one ionizable hydrogen, calculate the molar mass of the acid.
d. A 2.00 × 10−3 mole sample of pure acetylsalicylic acid was dissolved in mL of water and then titrated with M NaOH(aq). The equivalence point was reached after mL of the NaOH solution had been added. Using the data from the titration, shown in the table, determine
the value of the acid dissociation constant, Ka , for acetylsalicylic acid
2. the pH of the solution after a total volume of mL of the NaOH solution had
been added (assume that volumes are additive).

43. Note that the additional H+ from acetic acid was calculated in b
1970
a. What is the pH of a 2.0 molar solution of acetic acid. Ka acetic acid = 1.8 × 10–5?
Ka = 1.8 x = [H+][A-]/[HA] = x2/2.0
X = [H+] = 6.0 x 10-3
-log = pH = 2.22
b. A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter of a 1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the buffer solution.
CH3CO2H + NaOH  CH3CO2Na + H2O
initial 0.2 mol mol
final 0.1 mol mol mol
H+ = Ka = 1.8 x 10-5
c. Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to liter of the buffer prepared in (b). Compute the hydrogen ion concentration of the resulting solution.
Note 1/5 of volume from part b
CH3CO2H + NaOH  CH3CO2Na + H2O
initial 0.02 mol mol mol
0.03 mol H+/0.140 L = 0.214M
Note that the additional H+ from acetic acid was calculated in b
adding 0.05 mol strong acid will react with the CH3CO2-, but 0.03 mol of strong acid will remain: this will dominate the pH
c. Suppose that 0.10 liter of 0.50 molar sodium hydroxide is added to liter of the buffer prepared in (b). Compute the pH of the resulting solution.
Note 1/5 of volume from part b (again)
CH3CO2H + NaOH  CH3CO2Na + H2O
initial 0.02 mol mol mol
0.03 mol OH-/0.140 L = 0.214M
-log = pOH = 0.669;
pH = 13.33
adding 0.05 mol strong base will react with the CH3CO2H but 0.03 mol of strong base will remain: this will dominate the pH

44. 1977
The value of the ionization constant, Ka, for hypochlorous acid, HOCl, is 3.1 × 10–8.
Calculate the hydronium ion concentration of a molar solution of HOCl.
Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of
0.050 molar HOCl and molar sodium hypochlorite, NaOCl.
A solution is prepared by the disproportionation reaction below. Cl2 + H2O p HCl + HOCl
Calculate the pH of the solution if enough chlorine is added to water to make the concentration of HOCl equal to molar.
1977
The value of the ionization constant, Ka, for hypochlorous acid, HOCl, is 3.1 × 10–8.
a. Calculate the hydronium ion concentration of a molar solution of HOCl.
b. Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of molar HOCl and molar sodium hypochlorite, NaOCl.
c. A solution is prepared by the disproportionation reaction below. Cl2 + H2O  HCl + HOCl
Calculate the pH of the solution if enough chlorine is added to water to make the concentration of HOCl equal to molar.  
c. A solution is prepared by the disproportionation reaction below. Cl2 + H2O  HCl + HOCl
Calculate the pH of the solution if 7.09 mL of chlorine are added to enough water to make the volume of the solution 100 mL.

45. can solve either by Ka or pKa (H-H)…here is H-H
1977
The value of the ionization constant, Ka, for hypochlorous acid, HOCl, is 3.1 × 10–8.
Calculate the hydronium ion concentration of a molar solution of HOCl.
Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of
0.050 molar HOCl and molar sodium hypochlorite, NaOCl.
A solution is prepared by the disproportionation reaction below. Cl2 + H2O p HCl + HOCl
Calculate the pH of the solution if enough chlorine is added to water to make the concentration of HOCl equal to molar.
1977
The value of the ionization constant, Ka, for hypochlorous acid, HOCl, is 3.1 × 10–8.
a. Calculate the hydronium ion concentration of a molar solution of HOCl.
Ka = 3.1 x = [H+][A-]/[HA] = x2/0.05
X = [H+] = 4.0 x 10-5 M
b. Calculate the concentration of hydronium ion in a solution prepared by mixing equal volumes of molar HOCl and molar sodium hypochlorite, NaOCl.
can solve either by Ka or pKa (H-H)…here is H-H
pH = pKa + log [A-]/[HA]
= -log 3.1 x log 0.02/0.05
= [H3O+] = 7.76 x 10-8M
= – 0.397
= 7.11
c. A solution is prepared by the disproportionation reaction below. Cl2 + H2O  HCl + HOCl
Calculate the pH of the solution if enough chlorine is added to water to make the concentration of HOCl equal to molar.  
Reaction would also make .004M HCl which will dominate pH
[H3O+] = .004M; - log = pH = 2.40
c. A solution is prepared by the disproportionation reaction below. Cl2 + H2O  HCl + HOCl
Calculate the pH of the solution if 7.09 mL of chlorine are added to enough water to make the volume of the solution 100 mL.
0.1 mol HCl/0.1 L = 1M; will make 1.00M HCl
[H3O+] = 1.00M; - log 1 = pH = 0.00

46. 1982
A buffer solution contains 0.40 mole of formic acid, HCOOH, and 0.60 mole of sodium formate, HCOONa, in 1.00 liter of solution. The ionization constant, Ka, of formic acid is 1.8 × 10–4.
a. Calculate the pH of this solution.
pH = pKa + log [A-]/[HA]
= - log 1.8 x log 0.6/0.4 = = 3.92
b. If 100. milliliters of this buffer solution is diluted to a volume of 1.00 liter with pure water, the pH does not change. Discuss why the pH remains constant on dilution.
The pH does not change because the ratio of A- to HA remains constant.
c. A 5.00 milliliter sample of 1.00 molar HCl is added to 100. milliliters of the original buffer solution. Calculate the [H3O+] of the resulting solution.
Note that 1/10 of original buffer solution is used.
Ka method
H-H method
HCO2H D HCO2- + H+
Ka = [H+][A-]/HA
pH = pKa + log [A-]/[HA]
0.040 mol
0.060 mol
= log .055/.045
…add L x 1 mol/L = .005 mol H+
H+ = Ka[HA]/[A-]
= [H+] = 3.832
+ .005mol
-.005mol
= (1.8 x 10-4)(.045)/.055
= 1.47 x 10-4
.045mol
.055mol
= 1.47 x 10-4
d. A 800. milliliter sample of 2.00 molar formic acid is mixed with 200. milliliters of 4.80 molar NaOH. Calculate the [H3O+] of the resulting solution. Hint: All NaOH reacts.
H-H method
HCO2H + NaOH  HCO2Na + H2O
Ka method
1.6 mol
0.96 mol…all NaOH reacts
Ka = [H+][A-]/HA
pH = pKa + log [A-]/[HA]
0.64 mol
0 mol
0.96 mol
= log .96/.64
H+ = Ka[HA]/[A-]
= H+] = 3.832
= (1.8 x 10-4)(.64)/.96
= 1.2 x 10-4
= 1.2 x 10-4

47. 1982
A buffer solution contains 0.40 mole of formic acid, HCOOH, and 0.60 mole of sodium formate, HCOONa, in 1.00 liter of solution. The ionization constant, Ka, of formic acid is 1.8 × 10–4.
a. Calculate the pH of this solution.
b. If 100. milliliters of this buffer solution is diluted to a volume of 1.00 liter with pure water, the pH does not change. Discuss why the pH remains constant on dilution.
c. A 5.00 milliliter sample of 1.00 molar HCl is added to 100. milliliters of the original buffer solution. Calculate the [H3O+] of the resulting solution.
d. A 800. milliliter sample of 2.00 molar formic acid is mixed with 200. milliliters of 4.80 molar NaOH. Calculate the [H3O+] of the resulting solution. Hint: All NaOH reacts.