Question

What is the pH of your solution after titrating with 12.00mL; 0.0500 M NaOH? 6.00 mL of 0.100 M butanoic acid; Ka = 1.52 x 10-5

Answer 1

Given:

M(C3H7COOH) = 0.1 M

V(C3H7COOH) = 6 mL

M(NaOH) = 0.05 M

V(NaOH) = 12 mL

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

mol(C3H7COOH) = 0.1 M * 6 mL = 0.6 mmol

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

mol(NaOH) = 0.05 M * 12 mL = 0.6 mmol

We have:

mol(C3H7COOH) = 0.6 mmol

mol(NaOH) = 0.6 mmol

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

C3H7COO- here is strong base

C3H7COO- formed = 0.6 mmol

Volume of Solution = 6 + 12 = 18 mL

Kb of C3H7COO- = Kw/Ka = 1*10^-14/1.52*10^-5 = 6.579*10^-10

concentration ofC3H7COO-,c = 0.6 mmol/18 mL = 0.0333M

C3H7COO- dissociates as

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

0.0333 0 0

0.0333-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 ((6.579*10^-10)*3.333*10^-2) = 4.683*10^-6

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

so, x = 4.683*10^-6 M

[OH-] = x = 4.683*10^-6 M

use:

pOH = -log [OH-]

= -log (4.683*10^-6)

= 5.3295

use:

PH = 14 - pOH

= 14 - 5.3295

= 8.6705

Answer: 8.67