1. ap chemistry chapter 7. buffers, titrations, and solubility
problems need to be properly numbered, and typed in
keys need to be properly typed in
ap chemistry chapter 7. buffers, titrations, and solubility
need common ion problems and a solved problem in chapter
need half equivalence point problems.
Need to point out NH4+ is a weak base
need to point out typical pH of buffer systems (5is for acetic acid/acetate for example)
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
formulas:
Ka = [H+][A-]/[HA]
Kb = [OH-][HA]/[A-]
pH = pKa + log[A-]/[HA]
need a simple lab that shows buffering effect…also need lots of pH measuring experiments

2. buffers: do they behave as advertised?
Name________________________________________
buffers: do they behave as advertised?
For safety, all experiments must be approved prior to mixing
Substances available: acetic acid, sodium acetate, hydrochloric acid, sodium hydroxide, and phenolphthalein indicator.
Equipment available: Disposable pipettes, clear plastic cups.
Weak acids and their conjugate bases, when combined, are buffer systems meaning they resist a change in pH. Design two comparative experiments that demonstrate this and share it with the class.
Experiment chosen
results
explanation

3. Preparing Buffers and Buffer Capacity
 Introduction:: A buffer solution is one that is resistant to change in pH when small amounts of strong acid or base are added. For example, when 0.01 mole of strong acid or base are added to distilled water, the pH drops to 2 with the acid and rises to 12 with the base. If the same amount of acid or base is added to an acetic acid – sodium acetate buffer, the pH may only change a fraction of a unit.
Buffers are important in many areas of chemistry. When the pH must be controlled during the course of a reaction, the solutions are often buffered. This is often the case in biochemistry when enzymes or proteins are being studied. Our blood is buffered to a pH of Variations of a few tenths of a pH unit can cause illness or death. Acidosis is the condition when pH drops too low. Alkalosis results when the pH is higher than normal.
Two species are required in a buffer solution. One is capable of reacting with OH- and the other will react with H3O+. The two species must not react with each other. Many buffers are prepared by combining a weak acid and its conjugate (acetic acid and sodium acetate) or a weak base and its conjugate (ammonia and ammonium chloride). In general, the pH range in which a buffer solution is effective is +/- one pH unit on either side of the pKa. The Henderson–Hasselbalch provides the information needed to prepare a buffer.
There is a limit to the amount of acid or base that can be added to a buffer solution before one of the components is used up. This limit is called the buffer capacity and is defined as the moles of acid or base necessary to change the pH of one liter of solution by one unit.
Buffer Capacity = (number of moles of OH- or H3O+ added)
(pH change)(volume of buffer in L)
In this experiment, the Henderson-Hasselbalch equation will be used to determine the amount of acetic acid and sodium acetate required to prepare a series of buffer solutions. Once the buffer solutions have been prepared, their buffer capacity will be determined.
Purpose:The purpose of this experiment is to prepare buffer solutions and to determine their buffer capacity.
magnetic stirrer (if available)
100 mL volumetric flasks
 Equipment/Materials:
250 mL beaker
 acetic acid (0.10, 0.30, 0.50M)
stir bar
0.100 M NaOH
sodium acetate
Safety:
burets
 Always wear an apron and goggles in the lab.
buret clamp
Report any spills so they may be cleaned up.
pH meter
standard buffer solution (pH 4 & 7)

4. Procedure:
Before preparing the buffer solutions, you must determine the amount of acetic acid and sodium acetate required. The Ka of acetic acid is 1.8 X In the space below, show the calculations for the preparation of 100 mL of an acetic acid – sodium acetate pH 5.0 buffer using:
0.10 M acetic acid
0.30 M acetic acid
0.50 M acetic acid
1. Your lab group will be assigned to prepare one of the buffer solutions. Check your calculations with your instructor before proceeding.
2. The procedure for calibrating pH meters varies from instrument to instrument. Follow your instructor’s directions for this step.
3. Once the pH has been calibrated, measure the pH of the buffer solution that your group prepared. Record the value in your data table.
4. Clean two burets. Fill one with the buffer solution that you prepared and fill the other with the M NaOH. Make sure that the tips of the burets are filled and that the level of the liquids is at or below the 0.00 mL line.
5. Transfer mL of the buffer solution to a 250 mL beaker and add some distilled water. If a magnetic stirrer is to be used, keep the tip of the electrode above the stir bar.
6. Begin adding the NaOH to the buffer solution in small increments. After each addition, record the total volume of NaOH added and the pH of the solution in the data table.
7. Continue adding the NaOH solution until the pH has risen at least one pH unit.
8. Repeat as time allows.

5. Preparing Buffers and Buffer Capacity
Calculations:
Summarize the data for your titrations in the table below.
Show the calculations for one of the trials above.
What was the average buffer capacity for the buffer that your group prepared?
Questions:
How does the concentration of the buffer affect the buffer capacity?
What differences would be observed if HCl were used in place of NaOH?
Write equations to show how a buffer works.
Preparing Buffers and Buffer Capacity
Data:
Buffer Prepared: _____________________________
Concentration: _____________________________
pH of buffer solution _____________________________
Titration Data:
Trial 1: ml NaOH:_______ pH: _______
Trial 2: ml NaOH:_______ pH: _______
Trial 3: ml NaOH:_______ pH: _______
Trial 1
Trial 2
Trial 3
Vol of buffer, mL
Vol of buffer, L
Change in pH
Vol M NaOH
Moles of NaOH
Buffer Capacity

6. buffers and systems that contain both acid and base A buffer solution
(a) H3PO4 / KH2PO4
(b) NaClO4/HClO4
(c) C5H5N/C5H5NHCl
HCl/KCl
HCl/sodium acetate
why is this mixture special? Consider the effect of adding either acid or base
weak acid
conjugate base
yes
A buffer solution
resists a change in pH.
It is composed of
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. 1:
a weak acid or a weak base
example:
acetic acid (HC2H3O2)
and
conjugate base
strong acid no
The presence of a common ion suppresses the ionization of a weak acid or a weak base. 2:
it’s conjugate acid or base
example:
sodium acetate
consider the effect of adding adding acid to this mixture, noting that it has a lot of CH3CO2- in it
CH3COONa
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
this could be prepared a few different ways……
weak base
conjugate acid
yes
It keeps the acid or base from ionizing, rendering it relatively harmless.
how it works:
[acid]
[base]
change in pH no
strong acid
conjugate base
Add acid (H+)
increases
(in the form of HC2H3O2)
common ion
decreases
(reacts with acetate)
not much
a weak acid
strong acid
weak base
yes if excess acetate
CH3COOH (aq) + H2O D H3O+ (aq) + CH3COO- (aq)
and
yes if excess sodium acetate because it reacts to make the weak acid needed
H+ + CH3CO2-  CH3CO2H
CH3COO- (aq) + H2O D CH3CO2H (aq) + OH-
decreases
(and makes acetate)
a weak base
Add base (OH-)
increases
not much
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…
since the solution contains both a weak acid and a weak base, it will neutralize acids and bases
H+ + CH3CO2-  CH3CO2H
no because
HCl is a strong acid (complete dissociation, no equilibrium)
Cl- is a neutral ion (will not react with H+)
If we add a acid
H+ is neutralized by the weak base
H+ (aq) + CH3COO- (aq) D CH3COOH (aq)
…and if we add base:
OH- (aq) + CH3COOH (aq) D CH3COO- (aq) + H2O (l)
OH- neutralized by the weak acid
therefore:
buffer solutions resist a change in pH
Add strong base
Le Chateliers Principle could also be used as an explanation
Adding product prevents the forward reaction from occuring

7. Henderson-Hasselbalch equation
A rearrangement of the Ka equilibrium expression Ka = [H+][A-]/[HA]
pH = pKa + log ([A-] / [HA]) for acids
pH = pKa + log ([B] / [BH+]) for bases
For the pH to change by 1 unit, [A-] / [HA] must change by a factor of 10.
note that –log Ka = pKa
Use Henderson-Hasselbalch Equation to determine pH for 0.80M HF/0.50M NaF. Ka HF = 3.5 x 10-4
don’t have to derive it:
pH = pKa + log[A-]/[HA]
        = -log (3.5x10-4) + log[0.50]/[0.80]
        = (-0.20) = 3.26
Handy for buffer pH calculations

8. fastest: H-H solution fast: ice table method x = [H3O+] =2.1 x 10−5 M;
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
fastest: H-H solution
a “pH of the buffer system” problem
pH = pKa + log[A-]/[HA]
        = -log (1.8 x10-5) + log[0.60]/[0.70]
        = (-0.067) = 4.68
fast:
CH3CO2H(aq) D
H+(aq) +
CH3CO2-(aq)
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.600] [0.700]
ice table method
x = [H3O+] =2.1 x 10−5 M;
pH = −log (2.1 x 10−5) = 4.68
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
(as usual we can simplify this since 0.7 and 0.6 are so much larger than 10-5)
Compare to pH without buffer:
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = x [0.600] [0.700]
x = [H3O+] =2.1 x 10−5 M; pH = −log (2.1 x 10−5) = 4.68
1.8 x 10-5 = x2/0.7M;
x = [H+] = M;
pH = -log = 2.45…makes sense

9. What is the pH of the solution that results from adding 25. 0 mL of 0
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
normal
LH(aq) D
H+(aq) +
L-(aq)
initial
0.025
change -x +x
equilibrium
0.025-x =0.025 x
x = 0.025
fastest: H-H
Ka = 1.4 x 10-4 = H+ [L−] [LH] = x [0.025] [0.025] ;
x = [H3O+] = 1.4 x 10−4 M;
pH = −log (1.4 x 10−4) = 3.85
pH = pKa + log[A-]/[HA]
        = -log (1.4 x10-4) + log[0.025]/[0.025]
     = 3.85

10. Henderson-Hasselbalch:
How much sodium acetate (NaCH3CO2) must be added to 1.00L of 0.10M acetic acid (HCH3CO2) to give the solution of pH 4.50? Note that the Ka of acetic acid is 1.8 x 10-5.
traditional
Henderson-Hasselbalch:
[H+] = 10-pH = = 3.2 x 10-5M
pH = pKa + log[A-]/[HA]
        4.50 = -log (1.8 x10-5) + log[A-]/[0.025]
prefer traditional; use H-H for what is the pH questions
Ka = 1.8 x 10-5 = H+ [CH3CO2−] [CH3CO2H] = 3.2 x 10−5 M [x] [0.10]
x =[CH3CO2−] = 0.057M
0.057 mol 1.0 liter x mol NaCH3CO2 mol CH3CO2− x gol NaCH3CO2 mol NaCH3CO2
= 4.7 g
    = log [A-]/0.025
log[A-] =
4.50 = log[A-]/[0.025]
= log[A-]/[0.025]
log[A-] = x 10-4
[A-] = 0.057M

11. Henderson-Hasselbalch: for (a) traditional for (a)
8. A 1.000L buffer solution is composed of 1.360g of KH2PO4 and 5.677g of Na2HPO4. Note that the Ka of KH2PO4 is 6.2 x The molecular weight of KH2PO4 is g/mol and Na2HPO4 is g/mol.
What is the pH of the buffer solution?
Suggest a method to lower the pH of this buffer solution by 0.5 units.
Henderson-Hasselbalch:
for (a)
traditional for (a)
a. H2PO4- D HPO42- + H+; Ka = 6.2 x 10-8 = HPO42− [H+] [H2PO4−]
pH = pKa + log[A-]/[HA]
        pH = -log (6.2 x10-8) + log[5.677 g/141.96g/mol]/1.360 g/ g/mol]
= 7.81
(pH = log /.00999
= log 4.00 =
[H2PO4-] = [KH2PO4]
= g KH2PO4 L x mol KH2PO g KH2PO4 = M
[HPO42-] = [Na2HPO4]
= g Na2HPO4 L x mol Na2HPO g Na2HPO4 = M
5.6 x = [x] [ ] ;x = H+ = x 10−8;
pH = -log x 10-8 = 7.81
(b)
x = [H2PO4-] = [KH2PO4] = M
b. will lower the pH to 7.31 by adding KH2PO4.
[H+] = 10-pH = = x 10-8
molKH2PO4 l x g KH2PO4 mol KH2PO4
= 4.29 g KH2PO4
4.29 g KH2PO4 – g KH2PO4 =
add 2.94 g KH2PO4
6.2 x 10-8 = [4.898 x 10−8] [x]

12. 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) of known concentration.
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.
What happens in a titration:
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) the equivalence point
Procedure:
Slowly add base to unknown acid until the indicator changes color (pink)
monitor pH
alternative procedure:

13. titrations: four scenarios
pH at equivalence point
initial pH and how to measure
pH as titant is added before equivalence point…is buffer created?
pH after equivalence point
plot and example
low: = pH [H+] = - log [HX]
7: forms a neutral salt
strong acid strong base added
H+ consumed…slow increase
high
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+]
H+ consumed…buffer is created. Slow inc.
>7: forms a basic salt
high
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)
<7: forms a acidic salt
weak base strong acid added
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

14. Since [H+] = [A-] at the halfway point:
the halfway point for titrations
(also known as the half-equivalence point)
reaction is halfway done- note the volumes.
add pH = pKa problems
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
Since [H+] = [A-] at the halfway point:
pH = pKa + log 1 = pKa

15. pick a partner
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. Try to include the words “buret” , “titrant”, and “analyte” while making the 5 points below:
1. “our starting pH is ____ because___”
2. “our finishing pH is ____ because___”
3. “our endpoint pH is ____ because___”
4. “The halfway point is here and at this point ____________________ because_________________”
5. “An example that shows endpoint versus equivalence point would be ________”
draw in each for future years…pretty easy
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

16. 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.

17. 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 the interior of Io, one of Saturn's 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
The NASA scientist 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

18. 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
similarly,
Q < Ksp
Unsaturated solution
No precipitate
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
Precipitate will form
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
should include common ion effect

20. pH and Solubility
What is the pH of a saturated magnesium hydroxide solution, given that it has a Ksp of 1.2 x 10-11?
At pH less than 10.45
Lower [OH-]
Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
OH- (aq) + H+ (aq) H2O (l)
Increase solubility of Mg(OH)2
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
At pH greater than 10.45
Ksp = (s)(2s)2 = 4s3
Raise [OH-]
4s3 = 1.2 x 10-11
Decrease solubility of Mg(OH)2
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
pOH = pH = 10.45
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.

21. Ksp sample problems
1. Calculate the solubility of barium sulfate in moles per liter and grams per liter. Its Ksp is 1.1 x The molecular weight of BaSO4 is 233 g/mol.
4. 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)
note these are somewhat similar to first 6 questions of problem set
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-] = (4.9 x 10-3 )(4.9 x 10-3 ) = 2.4 x 10-5
2. Calculate the Ksp for CaF2 if the calcium ion concentration has been found to be 2.4 x 10-4M.
5. calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L, given that its Ksp is 2.2 x Comment on the solubility of each ion. The molar mass of Cu(OH)2 is g/mol
CaF2(s) D Ca2+(aq) + 2F-(aq)
Ksp = [Ca2+][F-]2 = (s)(2s)2
Ksp = (2.4 x 10-4)(4.8 x 10-4)2 = 5.5 x 10-11
Note: it is a bit easier to just do Ksp = 4s3 = 4(2.4 x 10-4)3
Cu(OH)2(s) D Cu2+(aq) + 2 OH-(aq)
Ksp = [Cu2+] ][OH-]2 = (s)(2s)2 = 4s3 = 2.2 x 10-20
3. Calculate the solubility of MgF2 in moles per liter and grams per liter. Its Ksp is 5.2 x 10-11
s = 3 ( 2.2 x 10 −20 4 ) = 1.8 x 10-5 mol/L
MgF2(s) D Mg2+(aq) + 2F-(aq)
Ksp 5.2 x = [Ca2+][F-]2 = (s)(2s)2 = 4s3
s = x 10− = 2.4 x 10-4M
2.4 x 10−4 mol L x g mol = g/L
1.8 x 10−5 mol Cu OH 2 L x g mol Cu OH 2 = 1.8 x 10−3 g/L
6. Why Ksp? Why not just have solubility tables?
goes beyond solubility of substances- shows concentration of each ion, and can quantify the common ion effect.

22. a different perspective on buffers and solubility equilibria: chemteam
Videos
Calculate Kp Given Initial Partial Pressures
Calculate Ksp from Molar Solubility I
Calculate Ksp from Molar Solubility II
Ksp: The Common Ion Effect I
Ksp: The Common Ion Effect II
Buffers Tutorials & Problem Sets
Introduction to Buffers
Buffers: The Henderson-Hasselbalch Equation
Buffers and Ksp Tutorials & Problem Sets
Henderson-Hasselbalch Equation
Using the Henderson-Hasselbalch Equation I
Using the Henderson-Hasselbalch Equation II
Using the Henderson-Hasselbalch Equation III
Using the Henderson-Hasselbalch Equation IV
Using the Henderson-Hasselbalch Equation V
Using the Henderson-Hasselbalch Equation VI
Using the Henderson-Hasselbalch Equation VII
Using the Henderson-Hasselbalch Equation VIII
Calculate the pKa
Calculate the pH of a buffer after some HCl is added
Ksp Tutorials
Writing Ksp Expressions
Solving Ksp Problems I: Calculating Molar Solubility Given the Ksp
Part One - s2
Part Two - 4s3
Part Three - 27s4
Part Four - 108s5
Two unusual molar solubility problems that didn't fit anywhere else
Solving Ksp Problems II: Calculating Ksp when
given the molar solubility
given the solubility in g/100mL
given the pH
given cell potential values
given titration data
given weight/volume (w/v) concentration data
Solving Ksp Problems III: Calculating Solubility in g/100mL, given the Ksp
Solving Ksp Problems IV: Acid Base-related
Calculate the pH when given the Ksp
What is the minimum pH required for precipitation?
Solving Ksp Problems V: Miscellaneous
Common Ion Effect (10)
Selective Precipitation
Three H-H problems with an acid and its salt
Three more H-H problems with an acid and its salt
Three H-H problems with a base and its salt
Three more H-H problems with a base and its salt
Some more H-H problems
Given the Ksp, Calculate pH
Given pH, Calculate Ksp
What is the minimum pH required for precipitation?
Here is the complete acid –base problem set
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23. Acid Base Problems & Videos
Ka related
Calculate the pH of a weak acid I
Calculate the pH of a weak acid II
Calculate the pH of a weak acid III
Calculate the pH of a weak acid IV
Calculate the pH of a weak acid V
Kb related
Calculate the pH of a weak base I
Calculate the pH of a weak base II
Calculate the pH of a weak base III
Salt related
Calculate the pH of salt of a weak acid
pH and pOH related
pH and pOH Calculations I
pH and pOH Calculations II
pH and pOH Calculations III
pH and pOH Calculations IV
pH and pOH Calculations V
Neutralization
Calculate the pH after neutralization I
Calculate the pH after neutralization II
Calculate the pH after neutralization III
Calculate the volume required for neutralization I
Calculate the volume required for neutralization II
Henderson-Hasselbalch Equation
Using the Henderson-Hasselbalch Equation I
Using the Henderson-Hasselbalch Equation II
Using the Henderson-Hasselbalch Equation III
Using the Henderson-Hasselbalch Equation IV
Using the Henderson-Hasselbalch Equation V
Using the Henderson-Hasselbalch Equation VI
Using the Henderson-Hasselbalch Equation VII
Using the Henderson-Hasselbalch Equation VIII
Calculate the pKa
Calculate the pH of a buffer after some HCl is added
Miscellaneous
Calculate the pH of a solution
Calculate the pH of two solutions after mixing
Calculate the hydroxide ion concentration
A trick pH calculation question
A fresh perspective on acids and bases: ChemTeam
This includes the last chapter
Problem Sets
Ka related
Solving Ka Problems: Part One
Solving Ka Problems: Part Two
Solving Ka Problems: Part Three
Given pH and molarity, calculate Ka
Given pH and other concentration data, calculate Ka
Given osmotic pressure data, calculate Ka and percent ionization (omit for ap)
Given thermodynamic data, calculate Ka (omit until thermo unit is complete)
Kb related
Solving Kb Problems: Part One
Solving Kb Problems: Part Two
Solving Kb Problems: Part Three
Given pH and molarity, Calculate Kb
Percent Dissociation related
Given pH and Ka, Calculate the Percent Dissociation
Given Concentration and Percent Dissociation, Calculate Ka
Given Concentration and Ka, Calculate the Percent Dissociation
Given Percent Dissociation, Calculate the Concentration
Ksp related
Given the Ksp, Calculate pH
Given pH, Calculate Ksp
What is the minimum pH required for precipitation?
Solutions of Salts
Calculations Involving Salts of Weak Acids
Calculations Involving Salts of Weak Bases
Given the Ka of an Acid, Calculate the pH of a Solution of the Salt of that Acid
Given the Kb of a Base, Calculate the pH of a Solution of the Salt of that Base
Given the pK of a Salt, Calculate the K of an Acid or a Base
The Henderson-Hasselbalch Equation
Three H-H problems with an acid and its salt
Three more H-H problems with an acid and its salt
Three H-H problems with a base and its salt
Three more H-H problems with a base and its salt
Some more H-H problems
Miscellaneous Problems
Titration problems (strong acids and bases)
Titration problems (weak acids and bases)
Miscellaneous problems
Acid Base Problems & Videos
Return to ChemTeam Main Menu Return to Acid Base Menu
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24. buffers
chapter
problems
Additional problems with solutions may be found in the KEY

25. 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. 25

26. 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.
b. solid ammonium chloride is added to 75 mL of 0.016M HCl
c. 20 g of NaCl is added to 1.0L of 0.10M sodium acetate
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

27. better change key for 7 and 8- questions were changed
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 2.2g 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
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.
better change key for 7 and 8- questions were changed

28. 9. 425 g of NaOH is dissolved in 2
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 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?

29. 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.
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.

30. 14. Calculate the pH in the titration of 25. 0 mL of 0
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.

31. 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.

32. 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?

33. 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)
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 milligrams 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?
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.
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. 33

34. 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 )
25. 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.

35. 26. Solid AgCl has been placed in a beaker of water
26. 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
27. 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

36. 28. Preparing a saturated solution of an ionic compound can readily accomplished by placing an excess of the salt in water and allowing the solution to saturate; the solution above the precipitated salt is a saturated solution. At 20 OC, a saturated aqueous solution of silver acetate, AgCH3CO2, contains 1.0 g of the silver compound dissoved in mL of solution. Calculate the Ksp for silver acetate.
30. Lead salts may be carcinogenic. Determine the solubility of lead (II)bromide(a) in moles per liter and in (b) grams per liter of pure water at 25 OC. Ksp PbBr2 = 2.40 x 10-5 at 25OC.
29. Gold can be dissolved in water by converting the atoms to ions. What is the molar concentration of Au+ in a saturated solution of gold(I)chloride? Ksp AuCl = 2.0 x
31. Calcium hydroxide dissolves in water to the extent of 1.78 g per liter. What is the Ksp for Ca(OH)2?

37. 32. (a)Determine the solubility of AgBr in moles per liter, given that it has a Ksp of 4.1 x (b) Compare this value with the molar solubility of AgBr in 225 mL to which 0.15 g of NaBr has been added.
34. Will a precipitate of Mg(OH)2 for when 25.0 mL of a M NaOH is combined with 75 mL of 0.10 M magnesium chloride?
33. Lead salts may be carcinogenic. Determine the solubility of lead (II)bromide(a) in moles per liter and in (b) grams per liter of pure water at 25 OC. Ksp PbBr2 = 2.40 x 10-5 at 25OC.
35. What is the solubility of calcium phosphate in water at 25 OC given that it has a Ksp value of 2.0 x at that temperature? How many micrograms of calcium phosphate will dissolve in 100 Liters of water?

38. 36
Why do we usually not quote the Ksp values for soluble ionic compounds?
39.
Using data from Table 17.2, calculate the molar solubility of calcium phosphate, which is a component of bones. 37
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? 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
40.
Using data from Table 17.2, calculate the solubility of CaF2 in g/L.

39. 43.
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.
41.
The pH of a saturated solution of a metal hydroxide MOH is Calculate the Ksp for the compound.
44.
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.
42.
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.)

40. 45.
Calculate the molar solubility of BaSO4 (a) in water and (b) in a solution containing 1.0 M  ions.