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Equilibrium
Solubility Product
10 questions from 10 papers, each with its answer.
- questions
- 10
- papers
- 10
- years asked, 2017–2026
- 3 of 10
- of Equilibrium
- 9.62%
Appears in
Concepts tested
- Ksp from pH 2
- Common ion effect and Ksp
- Ksp from pH of a saturated solution
- Ksp of AB2 type salt
- Ksp of M2S type salt
- Ksp of MX2 type salt
- ksp-silver-oxalate
- Order of precipitation
- solubility-product-ksp
All Questions
2019 1 question
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pH of a saturated solution of $Ca(OH)_2$ is 9. The solubility product ($K_{sp}$) of $Ca(OH)_2$ is:$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$ pH = 9, so pOH = 14 − 9 = 5 and $[OH^-]=10^{-5}\ M$ $[Ca^{2+}]=\frac{10^{-5}}{2}$ $K_{sp}=[Ca^{2+}][OH^-]^2=\frac{10^{-5}}{2}\times(10^{-5})^2=0.5\times10^{-15}$
2018 1 question
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The solubility of $BaSO_4$ in water is $2.42 \times 10^{-3}\ g L^{-1}$ at 298 K. The value of its solubility product ($K_{sp}$) will be (Given molar mass of $BaSO_4$ = 233 g $mol^{-1}$)
Convert g/L to mol/L, then $K_{sp} = s^2$ for a 1:1 salt.
Solubility of $BaSO_4$, $s = \frac{2.42\times10^{-3}}{233}\ (mol\ L^{-1}) = 1.04\times10^{-5}\ mol\ L^{-1}$ $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$, each with concentration s $K_{sp} = [Ba^{2+}][SO_4^{2-}] = s^2 = (1.04\times10^{-5})^2$ $= 1.08\times10^{-10}\ mol^2 L^{-2}$
2017 1 question
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Concentration of the $Ag^+$ ions in a saturated solution of $Ag_2C_2O_4$ is $2.2 \times 10^{-4}\ mol\ L^{-1}$. Solubility product of $Ag_2C_2O_4$ is
$[C_2O_4^{2-}] = [Ag^+]/2$; $K_{sp} = [Ag^+]^2[C_2O_4^{2-}]$.
$Ag_2C_2O_4(s) \rightleftharpoons 2Ag^+(aq) + C_2O_4^{2-}(aq)$, with concentrations 2s and s $K_{SP} = [Ag^+]^2[C_2O_4^{2-}]$ $[Ag^+] = 2.2\times10^{-4}$ M $\therefore [C_2O_4^{2-}] = \frac{2.2\times10^{-4}}{2}$ M $= 1.1\times10^{-4}$ M $\therefore K_{SP} = (2.2\times10^{-4})^2(1.1\times10^{-4}) = 5.324\times10^{-12}$
2015 AIPMT-I 1 question
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The $K_{sp}$ of $Ag_2CrO_4$, AgCl, AgBr and AgI are, respectively, $1.1\times10^{-12}$, $1.8\times10^{-10}$, $5.0\times10^{-13}$, $8.3\times10^{-17}$. Which one of the following salts will precipitate last if $AgNO_3$ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and $Na_2CrO_4$?The salt with the highest solubility precipitates last.
$Ag_2CrO_4$ ($A_2B$ type): $K_{sp} = 4s^3 = 1.1\times10^{-12} \Rightarrow s = (0.275\times10^{-12})^{1/3} \approx 6.5\times10^{-5}$ M
AgCl: $s = \sqrt{1.8\times10^{-10}} \approx 1.3\times10^{-5}$ M
AgBr: $s = \sqrt{5\times10^{-13}} \approx 7\times10^{-7}$ M
AgI: $s = \sqrt{8.3\times10^{-17}} \approx 9\times10^{-9}$ M
$Ag_2CrO_4$ is the most soluble, so it precipitates last.
2012 AIPMT-PRE 1 question
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pH of a saturated solution of $Ba(OH)_2$ is 12. The value of solubility product ($K_{sp}$) of $Ba(OH)_2$ ispH = 12, so pOH = 2 and $[OH^-] = 10^{-2}$ M
$Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$; if solubility is x, $[Ba^{2+}] = x$ and $[OH^-] = 2x$
$2x = 10^{-2} \Rightarrow x = 5\times10^{-3}$ M
$K_{sp} = [Ba^{2+}][OH^-]^2 = x(2x)^2 = 5\times10^{-3}\times(10^{-2})^2 = 5\times10^{-7}$
2011 AIPMT-MAINS 1 question
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In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $Ag^+$ and $Pb^{2+}$ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the $Cl^-$ concentration is 0.10 M. What will the concentrations of $Ag^+$ and $Pb^{2+}$ be at equilibrium? ($K_{sp}$ for AgCl = $1.8\times10^{-10}$, $K_{sp}$ for $PbCl_2$ = $1.7\times10^{-5}$)At equilibrium the solution is saturated with both chlorides and $[Cl^-] = 0.10$ M.
For AgCl: $K_{sp} = [Ag^+][Cl^-]$
$[Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8\times10^{-10}}{10^{-1}} = 1.8\times10^{-9}$ M
For $PbCl_2$: $K_{sp} = [Pb^{2+}][Cl^-]^2$
$[Pb^{2+}] = \frac{K_{sp}}{[Cl^-]^2} = \frac{1.7\times10^{-5}}{(10^{-1})^2} = \frac{1.7\times10^{-5}}{10^{-2}}$
$[Pb^{2+}] = 1.7\times10^{-3}$ M
So $[Ag^+] = 1.8\times10^{-9}$ M and $[Pb^{2+}] = 1.7\times10^{-3}$ M.
2010 AIPMT-PRE 1 question
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If pH of a saturated solution of $Ba(OH)_2$ is 12, the value of its $K_{sp}$ ispH = 12, so pOH = 14 − 12 = 2 and $[OH^-] = 10^{-2}$ M
$Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$; if the solubility is S, then $[Ba^{2+}] = S$ and $[OH^-] = 2S$
$2S = 10^{-2}$, so $S = \frac{10^{-2}}{2} = 5\times10^{-3}$ M
$K_{sp} = [Ba^{2+}][OH^-]^2$
$K_{sp} = (5\times10^{-3})(10^{-2})^2$
$K_{sp} = 5.00\times10^{-7}\ M^3$
2004 AIPMT 1 question
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The solubility product of a sparingly soluble salt $AX_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles/litre) is:-Let the solubility be s mol/L. $AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$
$[A^{2+}] = s$ and $[X^-] = 2s$
$K_{sp} = [A^{2+}][X^-]^2 = s(2s)^2 = 4s^3$
$4s^3 = 3.2 \times 10^{-11} \Rightarrow s^3 = 8 \times 10^{-12}$
$s = 2 \times 10^{-4}$ mol/L
2002 AIPMT 1 question
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Solubility of $MX_2$-type electrolytes is $0.5 \times 10^{-4}$ mole/lit. Then find out $K_{sp}$ of electrolytes : -$MX_2 \rightleftharpoons M^{2+} + 2X^-$. If the solubility is s, then $[M^{2+}] = s$ and $[X^-] = 2s$.
$K_{sp} = s(2s)^2 = 4s^3$
$s = 0.5 \times 10^{-4}$, so $s^3 = 0.125 \times 10^{-12} = 1.25 \times 10^{-13}$
$K_{sp} = 4 \times 1.25 \times 10^{-13} = 5 \times 10^{-13}$
2001 AIPMT 1 question
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Solubility of a $M_2S$ salt is $3.5 \times 10^{-6}$ then find out solubility product : -$M_2S \rightleftharpoons 2M^+ + S^{2-}$. If the solubility is s, then $[M^+] = 2s$ and $[S^{2-}] = s$.
$K_{sp} = (2s)^2(s) = 4s^3$
$K_{sp} = 4 \times (3.5 \times 10^{-6})^3 = 4 \times 42.875 \times 10^{-18}$
$K_{sp} = 1.715 \times 10^{-16} \approx 1.7 \times 10^{-16}$