Question

Consider the titration of a 40.0 mL sample of 0.150 M HCl with 0.200 M KOH

a) What is the initial pH?

b) What is the pH after the addition of 10.0 mL of the KOH?

c) What is the pH at the equivalence point?

d) What is the pH after 40.0 mL of KOH has been added?

Answer 1

a)when 0.0 mL of KOH is added
Given:
M(HCl) = 0.15 M
V(HCl) = 40 mL
M(KOH) = 0.2 M
V(KOH) = 0 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 40 mL = 6 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 0 mL = 0 mmol


We have:
mol(HCl) = 6 mmol
mol(KOH) = 0 mmol
0 mmol of both will react
remaining mol of HCl = 6 mmol
Total volume = 40.0 mL

[H+]= mol of acid remaining / volume
[H+] = 6 mmol/40.0 mL
= 0.15 M


use:
pH = -log [H+]
= -log (0.15)
= 0.8239
Answer: 0.824

b)when 10.0 mL of KOH is added
Given:
M(HCl) = 0.15 M
V(HCl) = 40 mL
M(KOH) = 0.2 M
V(KOH) = 10 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 40 mL = 6 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 10 mL = 2 mmol


We have:
mol(HCl) = 6 mmol
mol(KOH) = 2 mmol
2 mmol of both will react
remaining mol of HCl = 4 mmol
Total volume = 50.0 mL

[H+]= mol of acid remaining / volume
[H+] = 4 mmol/50.0 mL
= 8*10^-2 M


use:
pH = -log [H+]
= -log (8*10^-2)
= 1.0969
Answer: 1.10

c)
This is titration of strong acid and strong base
At equivalence point, equal mol of strong acid and strong base would react to form neutral and water
So, pH at equivalence point would be 7.00
Answer: 7.00

d)when 40.0 mL of KOH is added
Given:
M(HCl) = 0.15 M
V(HCl) = 40 mL
M(KOH) = 0.2 M
V(KOH) = 40 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 40 mL = 6 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 40 mL = 8 mmol


We have:
mol(HCl) = 6 mmol
mol(KOH) = 8 mmol
6 mmol of both will react

remaining mol of KOH = 2 mmol
Total volume = 80.0 mL

[OH-]= mol of base remaining / volume
[OH-] = 2 mmol/80.0 mL
= 2.5*10^-2 M


use:
pOH = -log [OH-]
= -log (2.5*10^-2)
= 1.6021


use:
PH = 14 - pOH
= 14 - 1.6021
= 12.3979
Answer: 12.40