Estimate Ksp for BaF2 based on a molar solubility of 3.610-3.
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Ksp of barium fluoride, BaF2 is 1.0 x 10^-6. Calculate its solubility in grams per liter, knowing that BaF2 has a molar mass of 208.23 g/mol: BaF2 (s) ⇌ Ba^2+ (aq) + 2F^- (aq) Ksp = [Ba^2+][F^-]^2 1.0 x 10^-6 = (x)(2x)^2 1.0 x 10^-6 = 4x^3 x^3 = 2.5 x 10^-7 x = (2.5 x 10^-7)^(1/3) x ≈ 6.3 x 10^-3 g/L
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A saturated solution of barium fluoride, $\mathrm{BaF}_{2}$, has $\left[\mathrm{Ba}^{2+}\right]=3.6 \times 10^{-3} \mathrm{M}$ and $[\mathrm{F}]=7.2 \times 10^{-3} \mathrm{M}$ What is the numerical value of $K_{\mathrm{sp}}$ for $\mathrm{BaF}_{2} ?$
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Calculate the molar solubility of BaF2 (Ksp=1.0x10^-6) in a 0.4 M solution of CsF.
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