In a titration experiment 5000 mL of 010 M acetic acid forma

In a titration experiment, 50.00 mL of 0.10 M acetic acid (formal concentration, Ka = 1.75*10^-5 at 25°C) was titrated with a 0.10 M naOH (formal concentration) at 25°C. The pH of the solution at 50.00 mL of this titrant will be

Solution

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^-14/1.75*10^-5 = 5.714*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 ((5.714*10^-10)*5*10^-2) = 5.345*10^-6

since c is much greater than x, our assumption is correct
so, x = 5.345*10^-6 M



[OH-] = x = 5.345*10^-6 M

use:
pOH = -log [OH-]
= -log (5.345*10^-6)
= 5.272


use:
PH = 14 - pOH
= 14 - 5.272
= 8.728
Answer: 8.73