how can you calculate the pH of(a) 0.300 M HONH2 (Kb = 1.1* 10^-8)(b) 0.300 M HONH3Cl(d) a mixture containing? | fedest.com

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how can you calculate the pH of(a) 0.300 M HONH2 (Kb = 1.1* 10^-8)(b) 0.300 M HONH3Cl(d) a mixture containing?

2007-11-02 07:18:34 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

1 answers

HONH2 + H2O <==> HONH3+ + OH-, Kb = 1.1*10^-8
Kb = 1.1*10^-8 = [HONH3+] *[OH-] /[HONH2] = [OH-]
pH = 14 + log([OH-])
= 14 + log(1.1*10^-8)
= 6.04

2007-11-05 15:18:19 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋

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