HF(aq) is a weak acid with pKs = 3.17. In a solution of HF(aq) the following...

HF(aq) is a weak acid with pKs = 3.17.

In a solution of HF(aq) the following equilibrium sets:

HF(aq)+F^- (aq) = HF₂^-(aq) K = 5.1

Calculate [F^-(aq)], [HF(aq)] and [HF₂^-(aq)] in a 0.10 M HF(aq) solution, where pH = 3.2

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queen_honey_blossom answered 1 year ago

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