The precipitate of M 2 S 3 obtained on mixing equal volumes of solutions S1 having - M 3- - -4- 10 -5 M and S2 having - S 2- - -2- 10 -3 M- Calculate its solubility product-

After mixing the equal volumes of two solutions, [M^3+]=12× 4 × 10^ 5M= 2 × 10^ 5M and [S^2 ]= 12× 2 × 10^ 3M= 1 × 10^ 3M .Now, M_2S_3 ⇌ 2M^3++ ...

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Standard XII

Chemistry

Solubility Product

The precipita...

Question

The precipitate of $M_{2} S_{3}$ obtained on mixing equal volumes of solutions $S_{1}$ having $[M^{3 +}] = 4 \times 10^{- 5} M$ and $S_{2}$ having $[S^{2 -}] = 2 \times 10^{- 3} M$ . Calculate its solubility product.

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Solution

After mixing the equal volumes of two solutions, $[M^{3 +}] = \frac{1}{2} \times 4 \times 10^{- 5} M = 2 \times 10^{- 5} M$ and $[S^{2 -}] = \frac{1}{2} \times 2 \times 10^{- 3} M = 1 \times 10^{- 3} M$ .
Now, $M_{2} S_{3} ⇌ 2 M^{3 +} + 3 S^{2 -}$
$∴ K_{s p}$ of $M_{2} S_{3} = [M^{3 +}]^{2} [S^{2 -}]^{3}$
$= (2 \times 10^{- 5})^{2} (1 \times 10^{- 3})^{3}$
$= 4 \times 10^{- 19}$

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The precipitate of M2S3 obtained on mixing equal volumes of solutions S1 having [M3+]=4×10−5M and S2 having [S2−]=2×10−3M. Calculate its solubility product.

After mixing the equal volumes of two solutions, [M3+]=12×4×10−5M=2×10−5M and [S2−]=12×2×10−3M=1×10−3M .Now, M2S3⇌2M3++3S2−∴Ksp of M2S3=[M3+]2[S2−]3 =(2×10−5)2(1×10−3)3 =4×10−19

The precipitate of $M_{2} S_{3}$ obtained on mixing equal volumes of solutions $S_{1}$ having $[M^{3 +}] = 4 \times 10^{- 5} M$ and $S_{2}$ having $[S^{2 -}] = 2 \times 10^{- 3} M$ . Calculate its solubility product.