Question

if the salt Baf2 in a 0.10M solution of NaF, how much should the solubility of Baf2 be then? compare the solubility to that in pure water?

Answer 1

Answer on question #74882

Since the solution is already 0.10 M in F-1 ions, we must make an addition to our equilibrium concentrations.

$$\mathrm{BaF2(aq)} = \mathrm{Ba} + 2\mathrm{(aq)} + 2\mathrm{F} - 1\mathrm{(aq)}$$ $$"x" \quad "2x + 0.10"$$

(at equilibrium)

$$\mathrm{Ksp} = [\mathrm{Ba} + 2][\mathrm{F} - 1]2$$

Because BaF2 is only slightly soluble, you might expect "2x" to be negligible compared to 0.10. In that case

$(2x + 0.10)(X)$ and substituting into the Ksp expression, we get $1.0 \times 10^{-6} = (x)(0.10)2$

solving for x, we get: $x = 1.0 \times 10^{-2} M$

the solubility of BaF2 is $1.0 \times 10^{-2} \mathrm{M}$ in NaF solution

the solubility of BaF2 in pure water is

$$\mathrm{Ksp} = [\mathrm{Ba} + 2][\mathrm{F} - 1]2$$ $$1.0 \times 10^{-6} = (x)(2x)2 = 4x3$$ $$\text{solving for } x, \text{ we get: } x = 6.30 \times 10^{-3} \mathrm{M}$$

Solubility of BaF2 in water is $6.30 \times 10^{-3} \mathrm{M}$

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