Answer to Question #333625 in Inorganic Chemistry for nashitaa

Question

How many grams of NaF would have to be added to 2.00 L of 0.100 M HF

to yield a solution with a pH = 4.00?

Answer 1

pH = pK_a + log ([F^-]/[HF])

K_a(HF) = 6.8 * 10^-4

pK_a = - log K_a = - log 6.8 * 10^-4 = 3.17

log ([F^-]/[HF]) = pH - pK_a

log ([F^-]/0.1М) = 4 - 3.17 = 0.83

[F^-]/0.1М = 10^0.83

[F^-] = 0.68 M

m(NaF) = 0.68 M * 2 L * 41.99 g/mol = 57.10 g

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