Question

You have 75.0 mL of 0.14 M HA, a weak acid. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the K, value of HA? Select one or more: a. 3.2 x 10-6 b.9.0 x 10-7 c.0.40 d. none of these e. 1.3 x 10-6

Answer 1

Given:
M(HA) = 0.14 M
V(HA) = 75 mL
M(NaOH) = 0.1 M
V(NaOH) = 30 mL


mol(HA) = M(HA) * V(HA)
mol(HA) = 0.14 M * 75 mL = 10.5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 30 mL = 3 mmol


We have:
mol(HA) = 10.5 mmol
mol(NaOH) = 3 mmol

3 mmol of both will react

excess HA remaining = 7.5 mmol
Volume of Solution = 75 + 30 = 105 mL
[HA] = 7.5 mmol/105 mL = 0.0714 M

[A-] = 3/105 = 0.0286 M

They form acidic buffer
acid is HA
conjugate base is A-

use:
pH = pKa + log {[conjugate base]/[acid]}
5.50 = pKa + log {0.0286/0.0714}
5.50 = pKa + log {0.401}
5.50 = pKa - 0.397
pKa = 5.897

Now use:
pKa = -log Ka
5.897 = -log Ka
Ka = 10^(-5.897)
Ka = 1.3*10^-6

Answer: e