1. Aim #13: How can we use a base (acid) to determine the concentration of an acid (base) (continued)?
H.W. # 13a
Study pp
Ans. ques. p. 753 # 60 (plot the titration curve),
65,69,74 (for problem 58 only), ,79

2. V A. Titrating a weak base with a strong acid
Problem: For the titration of mL of M NH3 with 0.10 M HCl, calculate the pH of the resulting solution a) after the addition of 0 mL HCl b) after the addition of mL HCl c) at the equivalence point d) after the addition of mL HCl
Ans: a) Kb = 1.8 x 10-5 = [NH4+][OH-] ≈ x2 ͘ [NH3]
x2 = (.050)(1.8 x 10-5) = 9.0 x [OH-] = x = 9.5 x 10-4
pOH = 3.02 and pH = 10.98

3. b) neutralization: (.050mmol/mL)(50.00 mL) = 2.5 mmol NH3
(0.10 mmol/mL)(10.00 mL) = 1.0 mmol HCl
mol NH3 remaining = 2.5 mmmol – 1.0 mmol = 1.5 mmol [NH3] = mmol = 2.5 x 10-2 M ( ) mL
[NH4+] = mmol = 1.7 x 10-2 M ( ) mL
equilibrium: NH3(aq) + H2O(ℓ) NH4+(aq) + OH-(aq)
Kb = 1.8 x 10-5 = [NH4+][OH-] = ( x)(x) [NH3] (.025 – x)
1.8 x 10-5 ≈ (.017)(x) and [OH-] = x = 2.6 x (.025) pOH = -log(2.6 x 10-5) = 4.58 and pH = 9.42

4. C) at the equivalence point we need 2
C) at the equivalence point we need 2.5 mmol HCl to neutralize the NH3 completely.
[NH4+] = 2.5 mmol = 3.3 x 10-2 M
equilibrium: NH4+(aq) + H2O(ℓ) NH3(aq) + H3O+(aq)
Ka = Kw = 1.0 x = 5.56 x = x ͘ Kb x x 10-2 – x
5.56 x ≈ x ͘ x 10-2
x2 = 1.85 x 10-11
[H+] = x = 4.3 x 10-6 M
pH = 5.37

5. neutralization: 5.0 mmool HCl – 2.5 mmol NH3 = 2.5 mmol HCl remaining
d) The [H+] is due to the excess of HCl (0.10 mmol/mL HCl)(50.00 mL) = 5.0 mmol HCl
neutralization: 5.0 mmool HCl – 2.5 mmol NH3 = 2.5 mmol HCl remaining
NH3(aq) + HCl(aq) → NH4+(aq) + Cl-(aq)
Because NH4+ is a weak acid and HCl is a strong acid
[H+] = [HCl] = mmol = 2.5 x 10-2 M ( )mL
pH = -log(2.5 x 10-2) = 1.60

6. B. Graphing the titration curve
14.00
10.00
7.00 pH
equivalence pt.
2.00
Volume of HCl added (mL)

7. VI Titrations of polyprotic acids – produce more than one equivalence point.
Problem: Sketch the curve of the titration of 25.0 mL 0.1 M H2CO with 0.10 M NaOH.
Ans: The polyprotic acid dissociates in 2 distinctive steps.
step 1: H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(ℓ) Ka1 = 4.3 x 10-7
step 2: HCO3-(aq) + OH-(aq) → CO32-(aq) + H2O(ℓ) Ka2 = 5.6 x 10-11

8. When VNaOH = ½ VH2CO3 , [H2CO3] = [HCO3-] and pH = pKa use 1st VNaOH = VH2CO3 , pH = 1st equivalence point equation When VNaOH = 3/2 VH2CO3 , [HCO3-] = [CO3-] and pH = pKa use 2nd VNaOH = 2VH2CO3 , pH = 2nd equivalence point equation
14.00
12.00
11.00
10.00 pH
7.00
6.00
3.00
0.00
Volume NaOH added (mL)

9. VII Acid-Base Indicators Endpoint = Equivalence Point Choose an indicator that minimizes the error
Acid-base indicators are themselves weak acids. (Acid is one color, its conjugate base is another.)
HIn(aq) H+(aq) + In-(aq) red blue
If Ka = [H+][In-] = 1.0 x [HIn]
Ka = [In-] If pH = 1.0 , [H+] [HIn]

10. Ka = 1.0 x 10-8 = 10-7 = 1 = [In-] ͘ [H+] 1.0 x 10-1 10,000,000 [HIn] the solution will be red.
As we add OH-, a color change will be noticeable when
[In-] = purple [HIn] 10
[H+] = Ka [HIn] = 1.0 x 10-8 (10) = 1.0 x [In-] (1)
OR pH = 7.00
As we add more OH-, pH ↑
If [In-] >> 1 , HI H+ + In- shifts to the right [HIn] the solution will be blue.

11. For [In-] = 1 ͘ [HIn] 10
pH = pKa + log(1/10) = pKa – 1
For [In-] = [HIn]
pH = pKa + log(10/1) = pKa + 1
Therefore, choose an indicator whose pKa equals pH (of the equivalence point) ± 1.