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# Answer to Question #163726 in Inorganic Chemistry for Mensah Abrampa

Question #163726

A solution of 0.15M NH3 is only 1.1% ionized in solution. Calculate the pH. The Kb of ammonia is 1.8*10^-5

1
2021-02-15T03:23:51-0500

The reaction is:-

"NH_{{3}_{(aq)}}+H_2O_{(l)}\\rightarrow NH^+_{4_{(aq})}+OH^-"

Given concentration of "NH_3=0.15M"

As It ionize 1.1% only So Concentration of "NH_4^+" will be-

"[NH_4^+]=\\dfrac{1.1\\times 0.15}{100}=1.65\\times 10^{-3}M"

Also, "k_b=\\dfrac{[NH_4^+][OH^-]}{[NH_3]}"

"1.8\\times 10^{-5}=\\dfrac{1.65\\times 10^{-3}[OH^-]}{.15}"

"\\Rightarrow [OH^-]=\\dfrac{0.27\\times 10^{-5}}{1.65\\times 10^{-3}}=1.63\\times 10^{-3}M"

"P_{OH}=-log([OH^-])=-log(1.63\\times 10^{-3})"

"=2.79"

As we know, "P_H+P_{OH}=14\\\\"

"P_H=14-P_{OH}=14-2.79=11.21"

Hence "P_H" is 11.21.

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