Question

Part A: How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M butanoic acid to the equivalence point? The Ka of butanoic acid is 1.5 x 10^-5.

Part B: What is the pH at the equivalence point?

Answer 1

1)

find the volume of NaOH used to reach equivalence point

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

0.0998 M *50.0 mL = 0.12M *V(NaOH)

V(NaOH) = 41.5833 mL

Answer: 41.58 mL

2)

Given:

M(C3H7COOH) = 0.0998 M

V(C3H7COOH) = 50 mL

M(NaOH) = 0.12 M

V(NaOH) = 41.5833 mL

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

mol(C3H7COOH) = 0.0998 M * 50 mL = 4.99 mmol

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

mol(NaOH) = 0.12 M * 41.5833 mL = 4.99 mmol

We have:

mol(C3H7COOH) = 4.99 mmol

mol(NaOH) = 4.99 mmol

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

C3H7COO- here is strong base

C3H7COO- formed = 4.99 mmol

Volume of Solution = 50 + 41.5833 = 91.5833 mL

Kb of C3H7COO- = Kw/Ka = 1*10^-14/1.5*10^-5 = 6.667*10^-10

concentration ofC3H7COO-,c = 4.99 mmol/91.5833 mL = 0.0545M

C3H7COO- dissociates as

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

0.0545 0 0

0.0545-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 ((6.667*10^-10)*5.449*10^-2) = 6.027*10^-6

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

so, x = 6.027*10^-6 M

[OH-] = x = 6.027*10^-6 M

use:

pOH = -log [OH-]

= -log (6.027*10^-6)

= 5.2199

use:

PH = 14 - pOH

= 14 - 5.2199

= 8.7801

Answer: 8.78