Given:
M(HCl) = 0.8 M
V(HCl) = 5 mL
M(NH3) = 0.4 M
V(NH3) = 50 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.8 M * 5 mL = 4 mmol
mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.4 M * 50 mL = 20 mmol
We have:
mol(HCl) = 4 mmol
mol(NH3) = 20 mmol
4 mmol of both will react
excess NH3 remaining = 16 mmol
Volume of Solution = 5 + 50 = 55 mL
[NH3] = 16 mmol/55 mL = 0.2909 M
[NH4+] = 4 mmol/55 mL = 0.0727 M
They form basic buffer
base is NH3
conjugate acid is NH4+
Kb = 1.77*10^-5
pKb = - log (Kb)
= - log(1.77*10^-5)
= 4.752
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.752+ log {7.273*10^-2/0.2909}
= 4.15
use:
PH = 14 - pOH
= 14 - 4.15
= 9.85
Answer: 9.85
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