Consider the titration of a 25.0 mL sample of 0.180 molL−1 CH3NH2 (Kb=4.4×10−4) with 0.155 molL−1 HBr.
Determine the pH at 5.0 mL of added acid
Given:
M(HBr) = 0.155 M
V(HBr) = 5 mL
M(CH3NH2) = 0.18 M
V(CH3NH2) = 25 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.155 M * 5 mL = 0.775 mmol
mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.18 M * 25 mL = 4.5 mmol
We have:
mol(HBr) = 0.775 mmol
mol(CH3NH2) = 4.5 mmol
0.775 mmol of both will react
excess CH3NH2 remaining = 3.725 mmol
Volume of Solution = 5 + 25 = 30 mL
[CH3NH2] = 3.725 mmol/30 mL = 0.1242 M
[CH3NH3+] = 0.775 mmol/30 mL = 0.0258 M
They form basic buffer
base is CH3NH2
conjugate acid is CH3NH3+
Kb = 4.4*10^-4
pKb = - log (Kb)
= - log(4.4*10^-4)
= 3.357
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.357+ log {2.583*10^-2/0.1242}
= 2.675
use:
PH = 14 - pOH
= 14 - 2.6747
= 11.3253
Answer: 11.33
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