C5H6NH3+ + H2O ----> C6H5NH2 + OH-
0.02 0 0 (initial)
0.02-x x x (at equilibrium)
Ka = 10^-14 / kb
= 10^-14 / (4.0*10^-10)
= 2.5*10^-5
Ka = [C6H5NH2] [OH-] /[C6H5NH3+]
2.5*10^-5 = x*x / (0.02-x)
since Ka is small, x will be small and it be ignored as compared to
0.02
above expression thus becomes,
2.5*10^-5 = x*x / (0.02)
x = 7.071*10^-4 M
so,
[OH-] = X = 7.071*10^-4 M
pOH = -log [OH-]
= -log (7.071*10^-4)
= 3.15
pH = 14 - pOH
= 14 - 3.15
= 10.85
Answer: 10.85
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