Question

Calculate the pH for 300. mL of a 0.850 M solution of the benzoic acid, (C6H5COOH), Ka=6.30x10-5 ) being titrated with 0.850 M NaOH at the following positions in the titration.
a) The initial pH (before any NaOH has been added). a. 1.436 b. 12.864 c. 5.435 d. 8.534 e. 2.136

b) The pH of the solution after 125.0 mL of 0.850 M NaOH has been added. a. 10.034 b. 4.054 c. 3.543 d. 8.534 e. 2.132

c) The pH of the solution at the equivalence point (i.e. after 300. mL of 0.850 M NaOH has been added). a. 8.914 b. 5.423 c. 7.000 d. 5.086 e. 2.365

d) The pH of the solution after 400. mL of 0.850 M NaOH has been added. a. 0.916 b. 4.578 c. 13.084 d. 4.201 e. 9.800

Answer 1

mmol of acid = MV = 300*0.85 = 255 mmol of acid...

a)

initially, there is only acid

so

HA <-> H+ + A-

Ka = [H+][A-]/[HA]

in equilbirium

[H+] = [A-] = x

[HA] = M-x = 0.85-x

substitute

6.3*10^-5 = x*x/(0.85-x)

x = H+ = 0.00728

pH = -log(0.00728) = 2.1378

b)

pH for V = 125 mL of base M = 0.85

mmol of base MV = 125*0.85 = 106.25

mmol of weak acid left = 255-106.25 = 148.75

mmol of conjugte formed = 0 + 106.25= 106.25

this is a buffer so

pH = pKa + log(A-/HA)

pH= -log(6.3*10^-5) + log(106.25/148.75) = 4.054

c)

in equivalce point

mmol of acid = mmol of base

255 = M*V

V = 255/0.85 = 300 mL

so total V = 300 + 300 = 600 mL

recaculate benzoate ion in solution

[A-] = 300*0.85/600 = 0.425

so

A- + H2O <-> HA + OH- Kb

hydolysis will occur

Kb = [HA][OH-]/[A-]

Kb = Kw/Ka = (10^-14)/(6.3*10^-5) =1.587*10^-10

1.587*10^-10 = x*x/(0.425-x)

x = OH = 8.21*10^-6

pOH = -log(8.21*10^-6

pOH = 5.0856

pH = 14-pOH = !4-5.0856 = 8.9144

d)

finally..

NaOH left over = 400*0.85 = 340 mmol of base

mmol of HA = 255

OH left = 340-255 = 85

Vtotal = 400+300 = 700

[OH-] =85/700 = 0.121428

´pH = 14 + log(0.121428) = 13.084