Question

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.

Answer 1

See Below

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.

`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