QUIZ CHAPTER # 15
ANSWER KEY
Student’s Name
1. Which of the following solutions will be a buffer solution?
a) KNO3 / HNO3
b) KI / NaI
c) NaF / HF
d) LiClO4 / HClO4
2. A carbonate buffer is prepared by dissolving NaHCO3 and Na2CO3 in water. What chemical
equation shows how this buffer neutralizes added strong base?
−
+
a) CO2−
3 (aq) + H3 O (aq) → HCO3 (aq) + H2 O(l)
𝟐−
−
b) 𝐇𝐂𝐎−
𝟑 (aq) + 𝐎𝐇 (aq) → 𝐂𝐎𝟑 (aq) + 𝐇𝟐 𝐎(l)
+
c) HCO−
3 (aq) + H3 O (aq) → H2CO3(aq) + H2 O(l)
d) H2CO3(aq) + OH − (aq) → HCO−
3 (aq) + H2 O(l)
3. What would be the pH of a solution containing 0.20 M hypochlorous acid, HOCl, and 0.90 M
potassium hypochlorite, KOCl ? The Ka value for hypochlorous acid, HOCl, is 3.5×10-8.
a) 8.11
[𝑂𝐶𝑙 − ]
0.90
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔
= − log(3.5 × 10−8 ) + 𝑙𝑜𝑔
= 7.46 + 0.65 = 8.11
[𝐻𝑂𝐶𝑙]
0.20
b) 7.46
c) 6.81
d) 7.81
1 of 2
4. What quantity (moles) of which reagent would you add to 1.0 L of the original buffer from
question 3 so that the resulting solution has pH = pKa?
𝑝𝐻 = 𝑝𝐾𝑎 , if [𝑂𝐶𝑙 − ] = [𝐻𝑂𝐶𝑙]
a) KOH, 0.35 moles
Since there is excess of [𝑂𝐶𝑙 − ], acid must be added:
b) HCl, 0.35 moles
−
𝑂𝐶𝑙(𝑎𝑞)
c) KOH, 0.70 moles
d) HCl, 0.70 moles
−
𝐻𝐶𝑙(𝑎𝑞) → 𝐻𝑂𝐶𝑙(𝑎𝑞) + 𝐶𝑙(𝑎𝑞)
+
𝑥
0.20
−𝑥
+𝑥
initial amount (moles): 0.90
change (moles): −𝑥
0.90 − 𝑥
after completion (moles):
0.20 + 𝑥
0
After HCl is added [𝑂𝐶𝑙 − ] = [𝐻𝑂𝐶𝑙] or 0.90 − 𝑥 = 0.20 + 𝑥,
𝑥 = 0.35 moles.
5. Consider the titration of 25.0 mL of 0.200 M solution of hydrochloric acid, HCl, by 0.100 M
solution of barium hydroxide, Ba(OH)2. Calculate the pH of the resulting solution after 20.0 mL
of Ba(OH)2 were added.
a) 3.00
mol HCl = 0.0250 L×0.200 M = 0.00500 moles
mol 𝐵𝑎(𝑂𝐻)2 = 0.0200 L×0.100 M = 0.00200 moles
b) 7.00
2 𝐻𝐶𝑙(𝑎𝑞) + 𝐵𝑎(𝑂𝐻)2(𝑎𝑞) → 𝐵𝑎𝐶𝑙2(𝑎𝑞) + 2 𝐻2 𝑂(𝑎𝑞)
c) 2.52
d) 1.65
[H + ] =
initial amount (moles): 0.00500
0.00200
change (moles): −0.00400
−0.00200
after completion (moles): 0.00100
0.00100 𝑚𝑜𝑙
= 2.2 × 10−2 𝑀
0.0450 𝐿
pH = −𝑙𝑜𝑔[H + ] = −log(2.2 × 10−2 ) = 1.65
2 of 2
0
0
+0.00200
0.00200