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If pKb for fluoride at 25°c is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is?(a) 3.52×10^-3(b) 6.75×10^-4(c) 5.38×10^-2(d) 1.74×10^-5 |
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Answer» The correct answer is (b) 6.75×10^-4 To explain: Kw = Ka × Kb Ka = Kw / Kb Ka = 10^-14/-log (10.83) = 6.75 × 10^-4. |
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