Question

The weak base hydroxylamine, HONH2, is titrated with HCl. What is the pH after 17 mL of 0.20 M HCl is added to 30.0 mL of 0.20 M HONH2? Kb of HONH2 = 1.1 x 10-8

Answer 1

Given:
M(HCl) = 0.2 M
V(HCl) = 17 mL
M(HONH2) = 0.2 M
V(HONH2) = 30 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.2 M * 17 mL = 3.4 mmol

mol(HONH2) = M(HONH2) * V(HONH2)
mol(HONH2) = 0.2 M * 30 mL = 6 mmol



We have:
mol(HCl) = 3.4 mmol
mol(HONH2) = 6 mmol

3.4 mmol of both will react
excess HONH2 remaining = 2.6 mmol
Volume of Solution = 17 + 30 = 47 mL
[HONH2] = 2.6 mmol/47 mL = 0.0553 M
[HONH3+] = 3.4 mmol/47 mL = 0.0723 M

They form basic buffer
base is HONH2
conjugate acid is HONH3+

Kb = 1.1*10^-8

pKb = - log (Kb)
= - log(1.1*10^-8)
= 7.9586

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 7.9586+ log {0.0723/0.0553}
= 8.0751

use:
PH = 14 - pOH
= 14 - 8.0751
= 5.92

PH = 5.92