A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 13.0 mL of H
Given:
M(HNO3) = 0.5 M
V(HNO3) = 13 mL
M(NH3) = 0.2 M
V(NH3) = 75 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.5 M * 13 mL = 6.5 mmol
mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.2 M * 75 mL = 15 mmol
We have:
mol(HNO3) = 6.5 mmol
mol(NH3) = 15 mmol
6.5 mmol of both will react
excess NH3 remaining = 8.5 mmol
Volume of Solution = 13 + 75 = 88 mL
[NH3] = 8.5 mmol/88 mL = 0.0966 M
[NH4+] = 6.5 mmol/88 mL = 0.0739 M
They form basic buffer
base is NH3
conjugate acid is NH4+
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {7.386*10^-2/9.659*10^-2}
= 4.628
use:
PH = 14 - pOH
= 14 - 4.6282
= 9.3718
Answer: 9.37
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