The K_b of ammonia (NH_3) is 1.8 times 10^{-5 | vedclass

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(A) For a weak base $NH_3$,the dissociation is $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$. The concentration of $OH^-$ ions is given by $[OH^-] =

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$11$

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(A) For a weak base $NH_3$,the dissociation is $NH_3 + H_2O ightleftharpoons NH_4^+ + OH^-$. The concentration of $OH^-$ ions is given by $[OH^-] = \sqrt{K_b 	imes C}$. $[OH^-] = \sqrt{1.8 	imes 10^{-5} 	imes 0.1} = \sqrt{1.8 	imes 10^{-6}} = 1.34 	imes 10^{-3} \ M$. $pOH = -\log[OH^-] = -\log(1.34 	imes 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.127 = 2.873$. Since $pH + pOH = 14$ at $298 \ K$,$pH = 14 - 2.873 = 11.127 \approx 11.13$.

The $K_b$ of ammonia $(NH_3)$ is $1.8 \times 10^{-5}$ at $298 \ K$. Calculate the $pH$ of a $0.1 \ M$ ammonia solution. (in $.13$)

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