Solubility product and the common ion effect — NEET & JEE Questions

Question 1 · difficulty L1 · recall

For a sparingly soluble salt $AB(s)\rightleftharpoons A^+ + B^-$, the solubility product is

• A. $K_{sp}=[A^+][B^-]$
• B. $K_{sp}=[AB]$
• C. $K_{sp}=[A^+]+[B^-]$
• D. $K_{sp}=1/[A^+][B^-]$

Question 2 · difficulty L2 · understanding

The $$\mathrm{pH}$$ at which $$\mathrm{Mg}(\mathrm{OH})_2\left[\mathrm{~K}_{\mathrm{sp}}=1 \times 10^{-11}\right]$$ begins to precipitate from a solution containing $$0.10 \mathrm{~M} \mathrm{~Mg}^{2+}$$ ions is __________.

Question 3 · difficulty L2 · understanding

$$25.0 \mathrm{~mL}$$ of $$0.050 ~\mathrm{M} ~\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$$ is mixed with $$25.0 \mathrm{~mL}$$ of $$0.020 ~\mathrm{M} ~\mathrm{NaF} . \mathrm{K}_{\mathrm{Sp}}$$ of $$\mathrm{BaF}_{2}$$ is $$0.5 \times 10^{-6}$$ at $$298 \mathrm{~K}$$. The ratio of $$\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}$$ and $$\mathrm{K}_{\mathrm{sp}}$$ is ___________. (Nearest integer)

Question 4 · difficulty L2 · understanding

At $298 \mathrm{~K}$, the solubility of silver chloride in water is $1.434 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$. The value of $-\log \mathrm{K}_{\mathrm{sp}}$ for silver chloride is _________. (Given mass of $\mathrm{Ag}$ is $107.9 \mathrm{~g} \mathrm{~mol}^{-1}$ and mass of $\mathrm{Cl}$ is $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Question 5 · difficulty L2 · understanding

If the solubility product of PbS is 8 $$\times$$ 10 $$-$$28 , then the solubility of PbS in pure water at 298 K is x $$\times$$ 10 $$-$$16 mol L $$-$$1 . The value of x is __________. (Nearest Integer) [Given : $$\sqrt2$$ = 1.41]

Question 6 · difficulty L2 · understanding

At $$310 \mathrm{~K}$$, the solubility of $$\mathrm{CaF}_{2}$$ in water is $$2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$$. The solubility product of $$\mathrm{CaF}_{2}$$ is ____________ $$\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}$$. (Give molar mass : $$\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$$)

Question 7 · difficulty L2 · understanding

Two salts A 2 X and MX have the same value of solubility product of 4.0 $$\times$$ 10 $$-$$12 . The ratio of their molar solubilities i.e. $${{S({A_2}X)} \over {S(MX)}}$$ = __________. (Round off to the Nearest Integer)

Question 8 · difficulty L2 · understanding

Arrange the following in increasing order of solubility product : $\mathrm{Ca}(\mathrm{OH})_2, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}$

• A. $\mathrm{PbS}<\mathrm{HgS}<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}$
• B. $\mathrm{HgS}<\mathrm{AgBr}<\mathrm{PbS}<\mathrm{Ca}(\mathrm{OH})_2$
• C. $\mathrm{HgS}<\mathrm{PbS}<\mathrm{AgBr}<\mathrm{Ca}(\mathrm{OH})_2$
• D. $\mathrm{Ca}(\mathrm{OH})_2<\mathrm{AgBr}<\mathrm{HgS}<\mathrm{PbS}$

Question 9 · difficulty L2 · understanding

$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?

• A. $\sqrt[5]{1.8 \times 10^{-30}}$
• B. $\frac{1.8 \times 10^{-30}}{27}$
• C. $\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$
• D. $\sqrt[2]{1.6 \times 10^{-30}}$

Question 10 · difficulty L2 · understanding

For a sparingly soluble salt $$\mathrm{AB}_2$$, the equilibrium concentrations of $$\mathrm{A}^{2+}$$ ions and $$B^{-}$$ ions are $$1.2 \times 10^{-4} \mathrm{M}$$ and $$0.24 \times 10^{-3} \mathrm{M}$$, respectively. The solubility product of $$\mathrm{AB}_2$$ is :

• A. $$0.069 \times 10^{-12}$$
• B. $$0.276 \times 10^{-12}$$
• C. $$6.91 \times 10^{-12}$$
• D. $$27.65 \times 10^{-12}$$