When 24.7 mL of a 0.500 M aqueous hydrofluoric acid solution is titrated with a 0.477 M aqueous potassium hydroxide solution....?

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When a 24.7 mL sample of a 0.500 M aqueous hydrofluoric acid solution is titrated with a 0.477 M aqueous potassium hydroxide solution, what is the pH at the midpoint in the titration?

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  • david
    Lv 7
    6 months ago

    0.0247 X 0.500 = 0.01235 moles HF

    midpoint -- 1/2 is neutralized == 0.006175 moles neutralized

    volume KOH = 0.006175 moles / 0.477 = 0.1295 liters

    total vol = 0.0247 + 0.1295 = 0.1542 L

    HF + KOH --> KF + H2O

    so 0.006175 moles NaF are formed

    HF <--> H+ + F- <<< HF is a weak acid so I need the Ka for HF

    ... the initial [HF] becomes 0.006175moles/0.1542 L = 0.40045 M

    initial [F-] = the same 0.40045 M

    initial [H+] = 0

    ==========================

    equilib == [HF] = 0.40045 - x

    .......... [F-] = 0.40045 + x

    ........... [H+] = x

    ===========================

    Ka = [H+][F-] / [HF] ... so look up Ka...

    Ka = x(0.40045 - x) / (0.40045 + x) <<< x is small .. round

    Ka = x(0.40045) / (0.40045)

    x = Ka ... = [H+] now just take -log to get pH ... so, look up Ka and finish

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