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5 December, 00:42

A 110.0 ml sample of 0.20 mhf is titrated with 0.10 mcsoh. determine the ph of the solution after the addition of 440.0 ml of csoh. the ka of hf is 3.5*10-4.

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  1. 5 December, 02:38
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    Given:

    Concentration of HF = 0.20 m

    Volume of HF = 110 ml

    Concentration of CsOH = 0.10 m

    Volume of CsOH = 440 ml

    Ka (HF) = 3.5 x 10^-4

    Balanced Chemical Equation:

    HF + CsOH = = = > CsF + H2O

    pH of the solution = log ([acid]/[base]) + pKa

    pH = log ([0.20 * (110/1000) ]/[0.10 * (440/1000) ]) - log (3.5x10^-4)

    pH = 3.15
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