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Pertanyaan

pH larutan dari 50 ml larutan NH3 0,2 m + 100 ml larutan NH4CL 0,1 m; kb NH3 = 1×10 pangkat -5

1 Jawaban

  • mol NH3 = 50×0,2 = 10 mmol
    mol NH4Cl = 100×0,1 = 10 mmol
    [OH-] = Kb × (mol NH3/mol NH4Cl) = 10^-5 × (10/10) = 10^-5 M
    pOH = - log 10^-5 = 5
    pH = 14 - 5 = 9

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