Calculate the pH of a buffer consisting of 50.00 mL of 0.12 M NH3 and 3.50 mL of 1.0 M HCl. (Given: Kb for NH3 = 1.76 x 10^(-5).
I'm mostly confused because it doesn't mention a conjugate acid for NH3, and HCl isn't its conjugate acid.
Given:
M(HCl) = 1 M
V(HCl) = 3.5 mL
M(NH3) = 0.12 M
V(NH3) = 50 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 1 M * 3.5 mL = 3.5 mmol
mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.12 M * 50 mL = 6 mmol
We have:
mol(HCl) = 3.5 mmol
mol(NH3) = 6 mmol
3.5 mmol of both will react
excess NH3 remaining = 2.5 mmol
Volume of Solution = 3.5 + 50 = 53.5 mL
[NH3] = 2.5 mmol/53.5 mL = 0.0467 M
[NH4+] = 3.5 mmol/53.5 mL = 0.0654 M
They form basic buffer
base is NH3
conjugate acid is NH4+
Kb = 1.76*10^-5
pKb = - log (Kb)
= - log(1.76*10^-5)
= 4.754
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.754+ log {6.542*10^-2/4.673*10^-2}
= 4.901
use:
PH = 14 - pOH
= 14 - 4.9006
= 9.0994
Answer: 9.10
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