# 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.

Apr 10, 2018

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#### Explanation:

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.

$C a {F}_{2}$
5.36gxx("1mole"/"78.07g") = "0.069moles"

$\left[{F}^{-} 1\right]$
$\frac{2 \times 0.069}{0.43 L}$ = 0.32 M

We need the ${K}_{B}$
${K}_{B} = {K}_{w} / {K}_{A} = \frac{1 \times {10}^{-} 14}{6.8 \times {10}^{-} 4} = 1.47 \times {10}^{-} 11$

${F}^{-} 1 + {H}_{2} O = H F + O {H}^{-} 1$

If you set up an ice table, you'll get:
${K}_{B} = \frac{{x}^{2}}{0.32 M}$ (assuming x<<0.32M)

x = $2.17 \times {10}^{-} 6 M O {H}^{-} 1$ If you add in the $1 \times {10}^{-} 7$ from water, you get:

pOH = -log($2.27 \times {10}^{-} 6 M$)
pOH = 5.64

pH = 14-pOH = 8.36