What is the pH of a solution made by dissolving 5.36 grams of calcium fluoride in enough water to make 430 mL of solution? The Ka for HF is 6.8x10–4.

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Answer 1 · 2018-04-10T03:16:27

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Pretty much the only thing you care about are those $F^-1$ ions in solution. They are a conjugate base and will give you a basic solution.

$CaF_2$
$5.36gxx("1mole"/"78.07g") = "0.069moles"$

$[F^-1]$
$(2xx0.069)/(0.43L)$ = 0.32 M

We need the $K_B$
$K_B = K_w/K_A = (1xx10^-14)/(6.8xx10^-4) = 1.47xx10^-11$

The equation we care about:
$F^-1 + H_2O = HF + OH^-1$

If you set up an ice table, you'll get:
$K_B = [x^2]/[0.32M]$ (assuming x<<0.32M)

x = $2.17xx10^-6 M OH^-1$ If you add in the $1xx10^-7$ from water, you get:

pOH = -log( $2.27xx10^-6 M$ )
pOH = 5.64

pH = 14-pOH = 8.36