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pH = pKNH4+ + log([NH3]/[NH4+]) = 9.23 + log([0.162]/[0.110]) = 9.23 + log[1.47] = 9.23 + 0.168 = 9.40

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Q: For a solution that is 0.162 m NH3 and 0.110 m NH4 cl calculate OH?
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