Engineer & Medical

Ionic Equilibrium question bank (Chemistry)

Practice 1,247 free Engineer & Medical Chemistry - Ionic Equilibrium questions with answers and step-by-step solutions.

Q1.

$M_3 A_2$ is a sparingly soluble salt of molar mass $y$ g mol$^{-1}$ and solubility $x$ g L$^{-1}$. The ratio of the molar concentration of the anion ($A^{3-}$) to the solubility product of the salt is

(A)
$\dfrac{1}{54} \cdot \dfrac{y^4}{x^4}$
(B)
$\dfrac{y^5}{108 x^4}$
(C)
$108 \cdot \dfrac{x^5}{y^5}$
(D)
$\dfrac{1}{108} \cdot \dfrac{y^4}{x^4}$
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Q2.

Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$. A solution containing $0.01 \mathrm{M} \mathrm{X}^{2+}$ and $0.01 \mathrm{M} \mathrm{Y}^{2+}$ is saturated with $\mathrm{H}_{2} \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$. (Nearest integer)
(Given: $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C}$, $\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_{2} \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30$, $\log 3=0.48, \log 5=0.70$)

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