Question

A chemist titrates 210.0 mL of a 0.7560 M butanoic acid (HC3H7CO2) solution with 0.6335 M KOH solution at 25 degrees celsius. Calculate the pH at equivalence. The pKa of butanoic acid is 4.82. Round your answer to 2 decimal places please

Answer 1

use:

pKa = -log Ka

4.82 = -log Ka

Ka = 1.514*10^-5

find the volume of NaOH used to reach equivalence point

M(HC3H7CO2)*V(HC3H7CO2) =M(NaOH)*V(NaOH)

0.756 M *210.0 mL = 0.6335M *V(NaOH)

V(NaOH) = 250.6077 mL

Given:

M(HC3H7CO2) = 0.756 M

V(HC3H7CO2) = 210 mL

M(NaOH) = 0.6335 M

V(NaOH) = 250.6077 mL

mol(HC3H7CO2) = M(HC3H7CO2) * V(HC3H7CO2)

mol(HC3H7CO2) = 0.756 M * 210 mL = 158.76 mmol

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

mol(NaOH) = 0.6335 M * 250.6077 mL = 158.76 mmol

We have:

mol(HC3H7CO2) = 158.76 mmol

mol(NaOH) = 158.76 mmol

158.76 mmol of both will react to form C3H7CO2- and H2O

C3H7CO2- here is strong base

C3H7CO2- formed = 158.76 mmol

Volume of Solution = 210 + 250.6077 = 460.6077 mL

Kb of C3H7CO2- = Kw/Ka = 1*10^-14/1.514*10^-5 = 6.607*10^-10

concentration ofC3H7CO2-,c = 158.76 mmol/460.6077 mL = 0.3447M

C3H7CO2- dissociates as

C3H7CO2- + H2O -----> HC3H7CO2 + OH-

0.3447 0 0

0.3447-x x x

Kb = [HC3H7CO2][OH-]/[C3H7CO2-]

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 ((6.607*10^-10)*0.3447) = 1.509*10^-5

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

so, x = 1.509*10^-5 M

[OH-] = x = 1.509*10^-5 M

use:

pOH = -log [OH-]

= -log (1.509*10^-5)

= 4.8213

use:

PH = 14 - pOH

= 14 - 4.8213

= 9.1787

Answer: 9.18