What is the pH of 0.1,M,N{H_3} | vedclass

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Question

$NH _3+ H _2 O ightleftharpoons NH _4+ OH ^{-}$

$Kb =\frac{\left[ NH _4^{+}ight]\left[ OH ^{-}ight]}{\left[ NH _3ight]}$

$1.8 	imes 10^

Vedclass · Accepted Answer

$11.13$

Vedclass · Answer

$NH _3+ H _2 O ightleftharpoons NH _4+ OH ^{-}$

$Kb =\frac{\left[ NH _4^{+}ight]\left[ OH ^{-}ight]}{\left[ NH _3ight]}$

$1.8 	imes 10^{-5}=\frac{ x ^2}{0.1}$

$\Rightarrow x =\sqrt{1.8 	imes 10^{-6}}$

$=1.34 	imes 10^{-3}$

$	herefore\left[ OH ^{-}ight]\left[ H ^{+}ight]=1.34 	imes 10^{-3}\left[ H ^{+}ight]=10^{-14}$

$\Rightarrow pH =11.13$

What is the $pH$ of $0.1\,M\,N{H_3}$

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