Question

Calculate the pH of a solution starting with 200.0 mL of 0.010 M butanoic acid (pKa = 4.818) solution that has been titrated to the equivalence point with 0.050 M NaOH. Ignore activities and state your answer with 3 sig. figs. Use approximations if you can.

Answer 1

use:

pKa = -log Ka

4.818 = -log Ka

Ka = 1.521*10^-5

find the volume of NaOH used to reach equivalence point

M(C3H7COOH)*V(C3H7COOH) =M(NaOH)*V(NaOH)

0.01 M *200.0 mL = 0.05M *V(NaOH)

V(NaOH) = 40 mL

Given:

M(C3H7COOH) = 0.01 M

V(C3H7COOH) = 200 mL

M(NaOH) = 0.05 M

V(NaOH) = 40 mL

mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)

mol(C3H7COOH) = 0.01 M * 200 mL = 2 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.05 M * 40 mL = 2 mmol

We have:

mol(C3H7COOH) = 2 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react to form C3H7COO- and H2O

C3H7COO- here is strong base

C3H7COO- formed = 2 mmol

Volume of Solution = 200 + 40 = 240 mL

Kb of C3H7COO- = Kw/Ka = 1*10^-14/1.521*10^-5 = 6.577*10^-10

concentration ofC3H7COO-,c = 2 mmol/240 mL = 0.0083M

C3H7COO- dissociates as

C3H7COO- + H2O -----> C3H7COOH + OH-

0.0083 0 0

0.0083-x x x

Kb = [C3H7COOH][OH-]/[C3H7COO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((6.577*10^-10)*8.333*10^-3) = 2.341*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.341*10^-6 M

[OH-] = x = 2.341*10^-6 M

use:

pOH = -log [OH-]

= -log (2.341*10^-6)

= 5.6306

use:

PH = 14 - pOH

= 14 - 5.6306

= 8.3694

Answer: 8.37