use:
pKb = -log Kb
9.37= -log Kb
Kb = 4.266*10^-10
Given:
M(HNO3) = 0.86 M
V(HNO3) = 30.4 mL
M(C6H5NH2) = 0.83 M
V(C6H5NH2) = 68 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.86 M * 30.4 mL = 26.144 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.83 M * 68 mL = 56.44 mmol
We have:
mol(HNO3) = 26.144 mmol
mol(C6H5NH2) = 56.44 mmol
26.144 mmol of both will react
excess C6H5NH2 remaining = 30.296 mmol
Volume of Solution = 30.4 + 68 = 98.4 mL
[C6H5NH2] = 30.296 mmol/98.4 mL = 0.3079 M
[C6H5NH3+] = 26.144 mmol/98.4 mL = 0.2657 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.266*10^-10
pKb = - log (Kb)
= - log(4.266*10^-10)
= 9.37
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.37+ log {0.2657/0.3079}
= 9.306
use:
PH = 14 - pOH
= 14 - 9.306
= 4.694
Answer: 4.69
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