The pH of a 0.1 M solution of $NH_{4}OH$ (having $K_{b} = 1.0 imes 10^{- 5})$ is equal to

Question

The pH of a 0.1 M solution of $NH_{4}OH$ (having $K_{b} = 1.0 	imes 10^{- 5})$ is equal to

Dr. Abubacker · Accepted Answer

(c)
$\left\lbrack {OH}^{-} ightbrack = \sqrt{K_{b} 	imes C}$
$\ \ \ \ \ \ \ \ \ \ \ \ \  = \sqrt{1 	imes 10^{- 5} 	imes 10^{- 1}}$
$= \sqrt{10^{- 6}} = 10^{- 3}$
$K_{w} = \left\lbrack H^{+} ightbrack\lbrack{OH}^{-}brack$
$10^{- 14} = \left\lbrack H^{+} ightbrack\lbrack 10^{- 3}brack$
$\left\lbrack H^{+} ightbrack = 10^{- 11}$
Hence, $pH = - \log H^{+}$
$= - \log{(1 	imes 10^{- 11})} = 11$

Q. The pH of a 0.1 M solution of $NH_{4}OH$ (having $K_{b} = 1.0 \times 10^{- 5})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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