Name ____Mr. Perfect_______________________________________ Date ______F 16___________
1. A 25.0 mL aliquot of 0.25 M HCNO (weak acid) is titrated with 0.15 M NaOH (strong base).
Calculate the pH after the following additions of the base solution: (10 pts)
{Problems are on both the front and back pages.} Ka = 2 x 10-4
Hint: round moles to 3 significant figures and report in scientific notation.
a) 0.00 mL of base (Before Addition)
HCNO ⇄ CNO- + H+
I 0.25
0
0
C –x
+x
+x
E 0.25-x
x
x
2.0 𝑥 10−4 =
𝑥2
𝑥2
≈
0.25 − 𝑥 0.25
x = 7.07 x 10-3 = [H+]
𝐶ℎ𝑒𝑐𝑘:
7.07 𝑥 10−3
× 100 = 2.83 % 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑖𝑠 𝑣𝑎𝑙𝑖𝑑
0.25
pH = -log (7.07 x 10-3) = 2.15
b) 20.00 mL of base. (before equivalence point)
0.020 L x 0.150 mol/L = 3.00 x 10-3 moles of base
0.025 L x 0.250 mol/L = 6.25 x 10-3 moles of acid
(6.25 x 10-3 mol acid - 3.00 x 10-3 mol base) = 3.25 x 10-3 moles acid remaining
[𝐻𝐴] =
3.25𝑥10−3 𝑚𝑜𝑙
= 0.0722 𝑀
0.045 𝐿
[𝐴− ] =
3.0𝑥10−3 𝑚𝑜𝑙
= 0.0667 𝑀
0.045 𝐿
Can use the Henderson-Hasselbach equation:
𝑝𝐻 = − log(2.0𝑥10−4 ) + 𝑙𝑜𝑔
[0.0667]
= 3.70 − 0.34 = 𝟑. 𝟔𝟕
[0.0722]
Chemistry 102 Quiz 5
Name ____Mr. Perfect_______________________________________ Date ______F 16___________
c) 41.67 mL of base. (at equivalence point)
NaOH:
0.04167 L x 0.15 mol/L = 6.25 x 10-3 mol base
molbase = molacid
*All of the weak acid HC6H5O has been converted to the conjugate base CNO-.
Total Volume = 25.00 mL + 41.67 mL = 66.67 mL or 0.06667 L
[𝐶𝑁𝑂 − ] =
6.25 𝑥 10−3 𝑚𝑜𝑙
= 0.0937 𝑀
0.06667 𝐿
CNO - + H2O ⇄ HCNO + OHI 0.0937
0
0
C -x
+x
+x
E 0.0937-x
x
x
𝐾𝑤
1 𝑥 10−14
𝐾𝑏 =
=
= 5.0 𝑥 10−11
𝐾𝑎 2.0 𝑥 10−4
5.0𝑥 10−11 =
𝑥2
𝑥2
≈
0.0937 − 𝑥 0.0937
𝑥 = 2.16 𝑥 10−6 = [𝑂𝐻 − ]
Check
2.16 𝑥 10−6
× 100 = 0.0023 % 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑖𝑠 𝑣𝑎𝑙𝑖𝑑.
0.0937
pOH = -log (2.16 x 10-6) = 5.66
pH = 14 – 5.66 = 8.34
d) 55.00 mL of base. (after equivalence point)
0.055 L x 0.150 mol/L = 8.25 x 10-3 moles of base
(8.25 x 10-3 moles of base - 6.25 x 10-3 mol acid) = 2.0 x 10-3 moles base remaining
Total volume = 25 mL + 55 mL = 80 mL
[𝑂𝐻 − ] =
2.0𝑥10−3 𝑚𝑜𝑙
= 0.025 𝑀
0.080 𝐿
𝑝𝑂𝐻 = − log(0.025) = 1.60
pH = 14 – 1.60 = 12.40
Chemistry 102 Quiz 5