pH = pKa + log [conj.base]/[acid]
Let the concentration of the conjugate base = x
=> The concentration of the acid is also x (As the acid concentration is equivalent to that of the conjugate base)
for CH₃COOH=0.50 M and CH₃COO=0,25 M
pH = pKa + log [salt]/[acid]
pKa = -log Ka = -log 1.8x10-5 = 4.74
pH = 4.74 + log (0.50 M/1.0 M) = 4.74 + log 0.5 = 4.74 + (-0.3)
pH = 4.74 - 0.3
pH = 4.44
For0.20 M Na2CO3
pH=−log[H+]
=−log (1.54×10−12)
=11.8
for HNO₂
Kb = 2.17X10^-11 = x^2 / 0.148
x = [OH-] = 1.79X10^-6
pOH = 5.747
pH = 14.000 - pOH = 8.253
for HPO₄³⁻
= [HPO₄³⁻] [PO₄³⁻] [H +]/×10⁻³
=[PO₄³⁻]5.464×10⁻¹⁸
Learn more about conjugate base here=
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