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Calculate the pH of 0.01 M NH4Cl solutio...

Calculate the pH of 0.01 M `NH_4Cl` solution at `25^@C` . `K_b` for `NH_4OH=1.81xx10^(-5)`

Text Solution

Verified by Experts

The hydrolysis reaction is
`NH_4^(+)+H_2O hArr NH_4OH + H^+`
pH of this solution is
`pH=1/2pK_w-1/2pK_b-1/2logc`
`pK_w=-log (1xx10^(-14))=14`
`pK_b=-log (1.81xx10^(-5))`=4.74
log c = log 0.01 = log `(1xx10^(-2))` =-2
`therefore pH=1/2(14)-1/2 (4.74)-1/2(-2)`
=5.63
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