Question

7. A sample of 0.10 M C6H5COOH(aq) (benzoic acid) solution is titrated with 0.10 M NaOH(aq) solution. What is the pH of the solution at the equivalence point? Ka(C6H5COOH) = 6.46 x 10-5 8. A sample of 0.10 M C6H5CH2NH2(aq) (benzylamine) solution is titrated with 0.10 M HBr(aq) solution. What is the pH of the solution at the equivalence point? Kb(C6H5CH2NH2) = 2.24 x 10-5

Answer 1

7)
Since concentration of both acid and base are same. The volume required for both to reach equivalence point will be same
Let 1 mL of both acid and base are required
Given:
M(C6H5COOH) = 0.1 M
V(C6H5COOH) = 1 mL
M(NaOH) = 0.1 M
V(NaOH) = 1 mL


mol(C6H5COOH) = M(C6H5COOH) * V(C6H5COOH)
mol(C6H5COOH) = 0.1 M * 1 mL = 0.1 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 1 mL = 0.1 mmol


We have:
mol(C6H5COOH) = 0.1 mmol
mol(NaOH) = 0.1 mmol

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

C6H5COO- here is strong base
C6H5COO- formed = 0.1 mmol
Volume of Solution = 1 + 1 = 2 mL
Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.46*10^-5 = 1.548*10^-10
concentration ofC6H5COO-,c = 0.1 mmol/2 mL = 0.05M

C6H5COO- dissociates as

C6H5COO-        + H2O   ----->     C6H5COOH +   OH-
0.05                        0         0
0.05-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 ((1.548*10^-10)*5*10^-2) = 2.782*10^-6

since c is much greater than x, our assumption is correct
so, x = 2.782*10^-6 M



[OH-] = x = 2.782*10^-6 M

use:
pOH = -log [OH-]
= -log (2.782*10^-6)
= 5.5556


use:
PH = 14 - pOH
= 14 - 5.5556
= 8.4444
Answer: 8.44

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