Vidaara.orgClass 11 · Chemistry
CodeVID-C11-07-T2-01
Assignment — Ionic Equilibrium: Acids & Bases
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 Lewis acid is a species that:
- A.donates a proton
- B.accepts a proton
- C.accepts an electron pair
- D.donates an electron pair
2.
The value of $K_w$ at $298\ \text{K}$ is:
- A.$10^{-7}$
- B.$10^{-14}$
- C.$10^{14}$
- D.$7$
3.
The pH of pure water at $298\ \text{K}$ is:
- A.0
- B.1
- C.7
- D.14
4.
For a weak acid, $[H^+]$ is given by:
- A.$c$
- B.$\sqrt{K_a c}$
- C.$K_a c$
- D.$K_a/c$
5.
A solution of sodium acetate ($CH_3COONa$) is:
- A.acidic
- B.neutral
- C.basic
- D.amphoteric
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define conjugate acid–base pair with one example.
7.
Calculate the pH of $0.0001\ \text{M}$ NaOH.
8.
State Ostwald's dilution law and write the expression for degree of ionisation of a weak acid.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
The $K_a$ of formic acid is $1.8\times10^{-4}$. Find $[H^+]$ in a $0.1\ \text{M}$ solution.
10.
Explain, with equations, why an aqueous solution of $NH_4Cl$ is acidic.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Define pH and $K_w$. Show that $pH+pOH=14$ at $298\ \text{K}$, and calculate the pH of $0.05\ \text{M}$ $H_2SO_4$ (assume complete ionisation).
Answer Key
Section A — Multiple Choice Questions
- (C) accepts an electron pair
- (B) $10^{-14}$
- (C) 7
- (B) $\sqrt{K_a c}$
- (C) basic
Section B — Short Answer (2 marks)
- Two species that differ by a single proton form a conjugate acid–base pair; e.g. $HCl$ (acid) and $Cl^-$ (its conjugate base), or $NH_4^+$ (acid) and $NH_3$ (its conjugate base).
- $[OH^-]=10^{-4}$, so $pOH=4$ and $pH=14-4=10$.
- For a weak electrolyte the degree of ionisation increases on dilution. For a weak acid $K_a=\dfrac{c\alpha^2}{1-\alpha}\approx c\alpha^2$, so $\alpha=\sqrt{K_a/c}$.
Section C — Short Answer (3 marks)
- $[H^+]=\sqrt{K_a c}=\sqrt{1.8\times10^{-4}\times0.1}=\sqrt{1.8\times10^{-5}}=4.24\times10^{-3}\ \text{M}$.
- $NH_4Cl$ is the salt of a strong acid and weak base. It gives $NH_4^+$, which reacts with water: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$. The released $H^+$ makes the solution acidic ($pH<7$); $Cl^-$ does not hydrolyse.
Section D — Long Answer (5 marks)
- $pH=-\log[H^+]$ and $K_w=[H^+][OH^-]=10^{-14}$ at $298\ \text{K}$. Taking $-\log$ of $K_w$: $-\log[H^+]-\log[OH^-]=14$, i.e. $pH+pOH=14$. For $0.05\ \text{M}$ $H_2SO_4$, each mole gives 2 $H^+$, so $[H^+]=0.10\ \text{M}$; $pH=-\log(0.10)=1$.
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