Question

Consider the titration of the titration of 50.0 mL of 0.100 M acetic acid (HC2H2O2) with 0.100 M. The pka = 4.76. d. Determine the pH after 50.0 mL of titrant (NaOH) have been added. This is the equivalence point. All of the acid has been converted to its conjugate base, pH is determined by the equilibrium for the conjugate base

Answer 1

Given:
pKa = 4.76

use:
pKa = -log Ka
4.76 = -log Ka
Ka = 1.738*10^-5
Given:
M(CH3COOH) = 0.1 M
V(CH3COOH) = 50 mL
M(NaOH) = 0.1 M
V(NaOH) = 50 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.1 M * 50 mL = 5 mmol

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


We have:
mol(CH3COOH) = 5 mmol
mol(NaOH) = 5 mmol

5 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base
CH3COO- formed = 5 mmol
Volume of Solution = 50 + 50 = 100 mL
Kb of CH3COO- = Kw/Ka = 1*10^-14/1.738*10^-5 = 5.754*10^-10
concentration ofCH3COO-,c = 5 mmol/100 mL = 0.05M

CH3COO- dissociates as

CH3COO-        + H2O   ----->     CH3COOH +   OH-
0.05                        0         0
0.05-x                      x         x


Kb = [CH3COOH][OH-]/[CH3COO-]
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 ((5.754*10^-10)*5*10^-2) = 5.364*10^-6

since c is much greater than x, our assumption is correct
so, x = 5.364*10^-6 M



[OH-] = x = 5.364*10^-6 M

use:
pOH = -log [OH-]
= -log (5.364*10^-6)
= 5.2705


use:
PH = 14 - pOH
= 14 - 5.2705
= 8.7295
Answer: 8.73