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Standard XII

Chemistry

Solubility Product

The solubilit...

Question

The solubility product of $P b I_{2}$ is $8.0 \times 10^{- 9}$ . The solubility of lead iodide in 0.1 molar solution of lead nitrate is $x \times 10^{- 6} m o l / L$ . Given: $\sqrt{2} = 1.41$ . The value of $x$ is

141.00

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141.0

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141

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Solution

$P b I_{2} (s) ⇌ P b^{2 +} (a q) + 2 I^{-} (a q)$
$S + 0.1 2 S$
$K_{s p} (P b I_{2}) = 8 \times 10^{9}$
$K_{s p} = [P b^{2 +}] [I^{-}]^{2}$
$8 \times 10^{- 9} = (S + 0.1) (2 S)^{2}$
$\Rightarrow S^{2} = 2 \times 10^{- 8}$
$S = 1.414 \times 10^{- 4} m o l / L$
$= x \times 10^{- 6} m o l / L$
$∴ x = 141.4 \approx 141$

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Similar questions

Q. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants. Determine also the molarities of individual ions.
The solubility product constants are:
Silver chromate,

1.1 \times 10^{- 12}

Barium chromate,

1.2 \times 10^{- 10}

Ferric hydroxide,

1.0 \times 10^{- 38}

Lead dichloride,

1.6 \times 10^{- 5}

Mercurous iodide,

4.5 \times 10^{- 29}

${PbI}_{2}$ has a solubility product of $8.0 \times 10^{- 9}$ . Lead Iodide has a solubility in $0.1 molar$ Lead Nitrate is $x \times 10^{- 6} mol / L$ . What is the value of $X$ ? (Round off to the nearest integer)

Q. The solubility product of lead iodide is

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. Calculate its molar solubility in

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Q. The solubility of

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Q. The solubility product of

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7.47 \times 10^{- 9}

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The solubility product of PbI2 is 8.0×10−9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10−6 mol/L. Given: √2=1.41. The value of x is

PbI2(s)⇌Pb2+(aq)+2I−(aq) S+0.1 2S Ksp(PbI2)=8×109 Ksp=[Pb2+][I−]2 8×10−9=(S+0.1)(2S)2 ⇒S2=2×10−8 S=1.414×10−4 mol/L =x×10−6 mol/L ∴x=141.4≈141

The solubility product of $P b I_{2}$ is $8.0 \times 10^{- 9}$ . The solubility of lead iodide in 0.1 molar solution of lead nitrate is $x \times 10^{- 6} m o l / L$ . Given: $\sqrt{2} = 1.41$ . The value of $x$ is