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Vidaara.orgClass 11 · Chemistry
CodeVID-C11-07-T3-01
Assignment — Solubility Product & Common Ion Effect
Chapter: Equilibrium
Topic: Solubility Product & Common Ion Effect
Maximum Marks: 30
Time: 60 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
  • Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
A buffer solution resists changes in:
  • A.temperature
  • B.pressure
  • C.pH
  • D.volume
2.
When $[salt]=[acid]$ in an acidic buffer, the pH equals:
  • A.$0$
  • B.$7$
  • C.$pK_a$
  • D.$14-pK_a$
3.
For $AgCl$ of solubility $s$, $K_{sp}$ is:
  • A.$s$
  • B.$s^2$
  • C.$4s^3$
  • D.$27s^4$
4.
The common ion effect on a sparingly soluble salt:
  • A.increases its solubility
  • B.decreases its solubility
  • C.has no effect
  • D.changes its $K_{sp}$
5.
A precipitate forms when:
  • A.$Q_{sp}
  • B.$Q_{sp}=K_{sp}$
  • C.$Q_{sp}>K_{sp}$
  • D.$Q_{sp}=0$
Section B — Short Answer (2 marks) 3 × 2 = 6 marks
6.
Define solubility product $K_{sp}$ for the salt $Ag_2CrO_4$ and express it in terms of solubility $s$.
7.
What is a buffer solution? Give one example of an acidic buffer.
8.
State the common ion effect with one example.
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
9.
Calculate the pH of a buffer containing $0.25\ \text{M}$ $NH_4OH$ and $0.05\ \text{M}$ $NH_4Cl$. ($pK_b=4.74$.)
10.
The $K_{sp}$ of $Mg(OH)_2$ is $1.0\times10^{-11}$. Calculate its molar solubility.
Section D — Long Answer (5 marks) 1 × 5 = 5 marks
11.
Define solubility product and ionic product. Using the criterion $Q_{sp}$ vs $K_{sp}$, decide whether a precipitate forms when equal volumes of $0.0002\ \text{M}$ $AgNO_3$ and $0.0002\ \text{M}$ $NaCl$ are mixed ($K_{sp}$ of $AgCl=1.8\times10^{-10}$).

Answer Key

Section A — Multiple Choice Questions
  1. (C) pH
  2. (C) $pK_a$
  3. (B) $s^2$
  4. (B) decreases its solubility
  5. (C) $Q_{sp}>K_{sp}$
Section B — Short Answer (2 marks)
  1. $Ag_2CrO_4\rightleftharpoons 2Ag^+ + CrO_4^{2-}$, so $[Ag^+]=2s$, $[CrO_4^{2-}]=s$; $K_{sp}=(2s)^2(s)=4s^3$.
  2. A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Example: a mixture of acetic acid and sodium acetate ($CH_3COOH + CH_3COONa$).
  3. Addition of an ion already present in a weak-electrolyte equilibrium suppresses its ionisation. E.g. adding $NH_4Cl$ to $NH_4OH$ decreases the ionisation of $NH_4OH$ (lowers $[OH^-]$).
Section C — Short Answer (3 marks)
  1. $pOH=pK_b+\log\dfrac{[salt]}{[base]}=4.74+\log\dfrac{0.05}{0.25}=4.74+\log 0.2=4.74-0.70=4.04$; so $pH=14-4.04=9.96$.
  2. $Mg(OH)_2\rightleftharpoons Mg^{2+}+2OH^-$, $K_{sp}=4s^3=1.0\times10^{-11}$; $s^3=2.5\times10^{-12}$, so $s=\sqrt[3]{2.5\times10^{-12}}\approx1.36\times10^{-4}\ \text{mol L}^{-1}$.
Section D — Long Answer (5 marks)
  1. $K_{sp}$ is the product of ion concentrations (each raised to its stoichiometric power) in a saturated solution at equilibrium; $Q_{sp}$ (ionic product) uses the actual concentrations. On mixing equal volumes the concentrations halve: $[Ag^+]=[Cl^-]=1\times10^{-4}\ \text{M}$. $Q_{sp}=[Ag^+][Cl^-]=(10^{-4})(10^{-4})=10^{-8}$. Since $Q_{sp}=10^{-8}>K_{sp}=1.8\times10^{-10}$, a precipitate of $AgCl$ forms until $Q_{sp}=K_{sp}$.
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