Question

You titrate 50.0 mL of 0.100 M benzoic acid (C6H5COOH) with 0.250 M KOH. What is the pH of the final solution at the equivalence point? Ka of C6H5COOH = 6.3 x 10-5.

If all work could be shown I would appreciate it! I'm super confused.

Answer 1

find the volume of KOH used to reach equivalence point
M(C6H5COOH)*V(C6H5COOH) =M(KOH)*V(KOH)
0.1 M *50.0 mL = 0.25M *V(KOH)
V(KOH) = 20 mL
Given:
M(C6H5COOH) = 0.1 M
V(C6H5COOH) = 50 mL
M(KOH) = 0.25 M
V(KOH) = 20 mL


mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)
mol(C6H5COOH) = 0.1 M * 50 mL = 5 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.25 M * 20 mL = 5 mmol


We have:
mol(C6H5COOH) = 5 mmol
mol(KOH) = 5 mmol

5 mmol of both will react to form C6H5COO- and H2O

C6H5COO- here is strong base
C6H5COO- formed = 5 mmol
Volume of Solution = 50 + 20 = 70 mL
Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.3*10^-5 = 1.587*10^-10
concentration ofC6H5COO-,c = 5 mmol/70 mL = 0.0714M

C6H5COO- dissociates as

C6H5COO-        + H2O   ----->     C6H5COOH +   OH-
0.0714                        0         0
0.0714-x                      x         x


Kb = [C6H5COOH][OH-]/[C6H5COO-]
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 ((1.587*10^-10)*7.143*10^-2) = 3.367*10^-6

since c is much greater than x, our assumption is correct
so, x = 3.367*10^-6 M



[OH-] = x = 3.367*10^-6 M

use:
pOH = -log [OH-]
= -log (3.367*10^-6)
= 5.4727


use:
PH = 14 - pOH
= 14 - 5.4727
= 8.5273
Answer: 8.53