At 310 K- the solubility of CaF2 in water is 2-34 - 10-3g-100 mL- The solubility product of CaF2 is - 10-8 mol-L 3Given molar mass- CaF2 - 78 g mol-1

CaF_2 s⇌sCa^2+ + 2s2F^ K_sp = s(2s)^2 = 4s^3 Solubility(s) nbsp; nbsp;= 2.34 × 10^ 3gmol mL = 2.34 × 10^ 3× 1078molelit ∴ K_sp = 4 ×(3 × 10^ 4)^3 = 108 × 1 ...

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Standard XII

Chemistry

Solubility Product

At 310 K, the...

Question

At $310 K$ , the solubility of $C a F_{2}$ in water is $2.34 \times 10^{- 3} g / 100 m L$ . The solubility product of $C a F_{2}$ is $\times 10^{- 8} {(m o l / L)}^{3}$
(Given molar mass: $C a F_{2} = 78 g m o l^{- 1}$ )

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Solution

$C a F_{2} s ⇌ C a^{2 +} s + 2 F^{-} 2 s$
$K_{s p} = s (2 s)^{2}$
$= 4 s^{3}$
Solubility(s) $= 2.34 \times 10^{- 3} \frac{g}{m o l} m L$
$= \frac{2.34 \times 10^{- 3} \times 10}{78} \frac{m o l e}{l i t}$
$∴ K_{s p} = 4 \times (3 \times 10^{- 4})^{3}$
$= 108 \times 10^{- 12}$
$= 0.0108 \times 10^{- 8} {(\frac{m o l e}{l i t})}^{3}$
$∴ x \approx 0$

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At 310 K, the solubility of CaF2 in water is 2.34×10−3g/100 mL. The solubility product of CaF2 is ×10−8(mol/L)3 (Given molar mass: CaF2=78 g mol−1)

CaF2s⇌Ca2+s+2F−2s Ksp=s(2s)2 =4s3 Solubility(s) =2.34×10−3gmolmL =2.34×10−3×1078molelit ∴Ksp=4×(3×10−4)3 =108×10−12 =0.0108×10−8(molelit)3 ∴x≈0

At $310 K$ , the solubility of $C a F_{2}$ in water is $2.34 \times 10^{- 3} g / 100 m L$ . The solubility product of $C a F_{2}$ is $\times 10^{- 8} {(m o l / L)}^{3}$
(Given molar mass: $C a F_{2} = 78 g m o l^{- 1}$ )