Question

What is the pH of the titration solution when 25.0 mL of .200 M aqueous formic acid, HCO₂H, is titrated with 50.0 mL of 0.100 M aqueous potassium hydroxide, KOH? The K_a for formic acid is 1.8x10^-4.

A. 8.52

B. 7.00

C. 5.71

D. 8.29

E. 10.26

Answer 1

Given:
M(HCO2H) = 0.2 M
V(HCO2H) = 25 mL
M(KOH) = 0.1 M
V(KOH) = 50 mL


mol(HCO2H) = M(HCO2H) * V(HCO2H)
mol(HCO2H) = 0.2 M * 25 mL = 5 mmol

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


We have:
mol(HCO2H) = 5 mmol
mol(KOH) = 5 mmol

5 mmol of both will react to form HCO2- and H2O

HCO2- here is strong base
HCO2- formed = 5 mmol
Volume of Solution = 25 + 50 = 75 mL
Kb of HCO2- = Kw/Ka = 1*10^-14/1.8*10^-4 = 5.556*10^-11
concentration ofHCO2-,c = 5 mmol/75 mL = 0.0667M

HCO2- dissociates as

HCO2-        + H2O   ----->     HCO2H +   OH-
0.0667                        0         0
0.0667-x                      x         x


Kb = [HCO2H][OH-]/[HCO2-]
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 ((5.556*10^-11)*6.667*10^-2) = 1.925*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.925*10^-6 M



[OH-] = x = 1.925*10^-6 M

use:
pOH = -log [OH-]
= -log (1.925*10^-6)
= 5.71


use:
PH = 14 - pOH
= 14 - 5.71
= 8.29
Answer: D