Question

Please help me with all parts. Thanks

Consider the titration of 25.0 mL of 0.125MHCl with 0.100MKOH. Calculate the pH after the addition of each of the following volumes of base.

Part A: 3.0mL

Part B: 20mL

Part C: 65mL

Answer 1

A)
Given:
M(HCl) = 0.125 M
V(HCl) = 25 mL
M(KOH) = 0.1 M
V(KOH) = 3 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.125 M * 25 mL = 3.125 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 3 mL = 0.3 mmol


We have:
mol(HCl) = 3.125 mmol
mol(KOH) = 0.3 mmol
0.3 mmol of both will react
remaining mol of HCl = 2.825 mmol
Total volume = 28.0 mL

[H+]= mol of acid remaining / volume
[H+] = 2.825 mmol/28.0 mL
= 0.1009 M


use:
pH = -log [H+]
= -log (0.1009)
= 0.9961
Answer: 0.9961

B)
Given:
M(HCl) = 0.125 M
V(HCl) = 25 mL
M(KOH) = 0.1 M
V(KOH) = 20 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.125 M * 25 mL = 3.125 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 20 mL = 2 mmol


We have:
mol(HCl) = 3.125 mmol
mol(KOH) = 2 mmol
2 mmol of both will react
remaining mol of HCl = 1.125 mmol
Total volume = 45.0 mL

[H+]= mol of acid remaining / volume
[H+] = 1.125 mmol/45.0 mL
= 0.025 M


use:
pH = -log [H+]
= -log (2.5*10^-2)
= 1.6021
Answer: 1.6021

C)
Given:
M(HCl) = 0.125 M
V(HCl) = 25 mL
M(KOH) = 0.1 M
V(KOH) = 65 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.125 M * 25 mL = 3.125 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.1 M * 65 mL = 6.5 mmol


We have:
mol(HCl) = 3.125 mmol
mol(KOH) = 6.5 mmol
3.125 mmol of both will react

remaining mol of KOH = 3.375 mmol
Total volume = 90.0 mL

[OH-]= mol of base remaining / volume
[OH-] = 3.375 mmol/90.0 mL
= 0.0375 M


use:
pOH = -log [OH-]
= -log (3.75*10^-2)
= 1.426


use:
PH = 14 - pOH
= 14 - 1.426
= 12.574
Answer: 12.57