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 15.0 mL of HNO3.
Given:
M(HNO3) = 0.5 M
V(HNO3) = 15 mL
M(NH3) = 0.2 M
V(NH3) = 75 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.5 M * 15 mL = 7.5 mmol
mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.2 M * 75 mL = 15 mmol
We have:
mol(HNO3) = 7.5 mmol
mol(NH3) = 15 mmol
7.5 mmol of both will react
excess NH3 remaining = 7.5 mmol
Volume of Solution = 15 + 75 = 90 mL
[NH3] = 7.5 mmol/90 mL = 0.0833 M
[NH4+] = 7.5 mmol/90 mL = 0.0833 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 {8.333*10^-2/8.333*10^-2}
= 4.745
use:
PH = 14 - pOH
= 14 - 4.7447
= 9.2553
Answer: 9.26
Get Answers For Free
Most questions answered within 1 hours.