What mass of sodium fluoride must be added to 250. mL of a 0.100 M HF solution to give a buffer solution having a pH of 3.50?

NetherCraft 0

Assume adding the solid NaF does not change the volume of the solution.

(Ka(HF) = 7.1 × 10-4)

1 Answer

  • pH = 3.50 ;

    [H+] = 10^-3.5

    [H+] = 3.16 x 10^-4

    [H+][F-]/[HF] = Ka = 7.1 x 10^-4

    3.16 x 10^-4[F-]/0.10 = 7.1 x 10^-4

    [F-] = 0.10(7.1 x 10^-4)/(3.16 x 10^-4)

    [F-] = 0.225

    0.225 moles F-/liter x 250 mL x 1 liter/1000 mL = 0.0562 moles F-

    0.0562 moles F- x 1mole NaF/1mole F- x 42 grams NaF/1mole NaF= 2.36 g NaF

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