Multiple Choice Questions (MCQs) for JEE Chemistry: Ionic Equilibrium
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Multiple Choice Questions
(a) \( 2.22\times10^{-5} \)
(b) \( 2.02\times10^{5} \)
(c) \( 4.33\times10^{4} \)
(d) \( 3.03\times10^{-5} \)
Show Answer & Explanation
Answer: (a) \( 2.22\times10^{-5} \)
Explanation:
1. \( K_h=K_w/K_a \) for the salt of a weak acid.
2. \( K_h=10^{-14}/(4.5\times10^{-10})=2.22\times10^{-5} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) A is 10 times stronger than B
(b) Strength of A : strength of B = 4 : 5
(c) their strengths cannot be compared
(d) B is 10 times stronger than A
Show Answer & Explanation
Answer: (a) A is 10 times stronger than B
Explanation:
1. A smaller \( pK_a \) means a larger \( K_a \): \( K_a(A)=10^{-4} \) and \( K_a(B)=10^{-5} \).
2. The ratio of \( K_a \) values is 10, so A is 10 times stronger (in terms of \( K_a \)).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 1 : 4
(b) 4 : 1
(c) 1 : 16
(d) 16 : 1
Show Answer & Explanation
Answer: (b) 4 : 1
Explanation:
1. At the same concentration \( [H^+]\propto\sqrt{K_a} \).
2. Ratio = \( \sqrt{3.14\times10^{-4}/1.96\times10^{-5}}=\sqrt{16}=4 \), i.e. 4 : 1.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 1.4\times10^{-4} \)
(b) \( 8.4\times10^{-4} \)
(c) 0.84
(d) 1.4
Show Answer & Explanation
Answer: (b) \( 8.4\times10^{-4} \)
Explanation:
1. \( K_{sp}=4s^3 \), so \( s=(3\times10^{-12})^{1/3}=1.44\times10^{-4} \) mol \( L^{-1} \).
2. Molar mass = 58 g/mol, so 1 L holds \( 8.4\times10^{-3} \) g.
3. 100 \( cm^3 \) holds \( 8.4\times10^{-4} \) g.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 25%
(b) 50%
(c) 75%
(d) 85%
Show Answer & Explanation
Answer: (c) 75%
Explanation:
1. Isotonic means equal osmotic pressure: \( i\times0.004=0.010 \), so \( i=2.5 \).
2. For \( \ce{Na2SO4} \), \( i=1+2\alpha \), giving \( \alpha=0.75 \), i.e. 75%.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 1\times10^{-8} \)
(b) \( 8\times10^{-8} \)
(c) \( 1.1\times10^{-8} \)
(d) \( 0.18\times10^{-8} \)
Show Answer & Explanation
Answer: (a) \( 1\times10^{-8} \)
Explanation:
1. \( \ce{Cr(OH)3} \) gives \( s \) of \( \ce{Cr^3+} \) and \( 3s \) of \( \ce{OH-} \): \( K_{sp}=27s^4 \).
2. \( s=(2.7\times10^{-31}/27)^{1/4}=(10^{-32})^{1/4}=10^{-8} \) mol \( L^{-1} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 5.63
(b) 4.74
(c) 8.37
(d) 9.37
Show Answer & Explanation
Answer: (c) 8.37
Explanation:
1. \( \ce{CH3COONa} \) is the salt of a weak acid and a strong base.
2. \( pH=7+\tfrac12(pK_a+\log C)=7+\tfrac12(4.74-2)=8.37 \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) S of AgBr is less than that of AgCl
(b) S of AgBr is greater than that of AgCl
(c) S of AgBr is equal to that of AgCl
(d) S of AgBr is \( 10^6 \) times that of AgCl
Show Answer & Explanation
Answer: (a) S of AgBr is less than that of AgCl
Explanation:
1. Both salts are of the same type, so \( S=\sqrt{K_{sp}} \).
2. \( S_{AgCl}=1.1\times10^{-5} \) and \( S_{AgBr}=5.9\times10^{-7} \), so AgBr is less soluble.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 2\times10^{-3} \)
(b) \( 4\times10^{-4} \)
(c) \( 1.6\times10^{-5} \)
(d) \( 1.8\times10^{-5} \)
Show Answer & Explanation
Answer: (a) \( 2\times10^{-3} \)
Explanation:
1. \( K_{sp}=4s^3 \).
2. \( s=(3.2\times10^{-8}/4)^{1/3}=(8\times10^{-9})^{1/3}=2\times10^{-3} \) mol \( L^{-1} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 3.0\times10^{-8}\ M \)
(b) \( 2.5\times10^{-7}\ M \)
(c) \( 2.8\times10^{-9}\ M \)
(d) \( 2.5\times10^{7}\ M \)
Show Answer & Explanation
Answer: (c) \( 2.8\times10^{-9}\ M \)
Explanation:
1. The common ion fixes \( [Cl^-]\approx0.1 \) M.
2. \( s=[Ag^+]=K_{sp}/[Cl^-]=2.8\times10^{-10}/0.1=2.8\times10^{-9} \) M.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 4\times10^{-5} \)
(b) \( 2\times10^{-5} \)
(c) \( 4\times10^{-10} \)
(d) \( 16\times10^{-20} \)
Show Answer & Explanation
Answer: (b) \( 2\times10^{-5} \)
Explanation:
1. For a binary salt \( K_{sp}=s^2 \).
2. \( s=\sqrt{4\times10^{-10}}=2\times10^{-5} \) mol \( dm^{-3} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 1\times10^{-4} \)
(b) \( 1\times10^{-2} \)
(c) \( 1\times10^{-3} \)
(d) \( 1\times10^{-5} \)
Show Answer & Explanation
Answer: (b) \( 1\times10^{-2} \)
Explanation:
1. \( K_h=10^{-14}/10^{-5}=10^{-9} \).
2. \( h=\sqrt{10^{-9}/0.1}=10^{-4} \), so percentage hydrolysis = \( 10^{-2} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 1.96\times10^{-2}\ mol/L \)
(b) \( 1.96\times10^{-3}\ mol/L \)
(c) \( 1.5\times10^{-4}\ mol/L \)
(d) \( 1.96\times10^{-1}\ mol/L \)
Show Answer & Explanation
Answer: (c) \( 1.5\times10^{-4}\ mol/L \)
Explanation:
1. \( [H^+]=10^{-3.82}=1.5\times10^{-4} \) mol/L.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 1.82\times10^{-5} \)
(b) \( 18.2\times10^{-5} \)
(c) \( 0.182\times10^{-5} \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( 1.82\times10^{-5} \)
Explanation:
1. \( K_a=C\alpha^2/(1-\alpha)=0.1\times(0.0134)^2/0.9866 \).
2. \( K_a=1.82\times10^{-5} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 3.42\times10^{-6} \)
(b) \( 3.42\times10^{-8} \)
(c) \( 1.00\times10^{-14} \)
(d) \( 2.00\times10^{-16} \)
Show Answer & Explanation
Answer: (d) \( 2.00\times10^{-16} \)
Explanation:
1. \( [H_2O]=55.5 \) M and \( \alpha=1.9\times10^{-9} \), so \( [H^+]=[OH^-]\approx10^{-7} \) M.
2. \( K=10^{-14}/55.5=1.8\times10^{-16}\approx2\times10^{-16} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 8
(b) 7
(c) above 8
(d) below 7
Show Answer & Explanation
Answer: (d) below 7
Explanation:
1. An acid can never make the solution basic.
2. \( \ce{H+} \) from water (\( 10^{-7} \) M) adds to \( 10^{-8} \) M, giving about \( 1.05\times10^{-7} \) M, so pH is about 6.98, which is below 7.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 0
(b) 1
(c) more than 7
(d) 7
Show Answer & Explanation
Answer: (c) more than 7
Explanation:
1. NaOH = 10 x 0.1 = 1 meq; \( \ce{H2SO4} \) = 10 x 0.05 = 0.5 meq.
2. 0.5 meq of NaOH is left over, so the solution is basic (pH > 7).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 3
(b) 8
(c) 11
(d) 14
Show Answer & Explanation
Answer: (b) 8
Explanation:
1. With equal concentrations of acid and conjugate base, \( pH=pK_a \).
2. \( pK_a=-\log10^{-8}=8 \), so pH = 8.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 7.302
(b) 9.302
(c) 8.302
(d) 10.302
Show Answer & Explanation
Answer: (c) 8.302
Explanation:
1. \( pK_a=-\log(5\times10^{-10})=9.302 \).
2. \( pH=pK_a+\log([salt]/[acid])=9.302+\log(0.15/1.5)=9.302-1=8.302 \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 3.4
(b) 2.6
(c) 1.8
(d) 0.56
Show Answer & Explanation
Answer: (c) 1.8
Explanation:
1. pOH = 14 - 9 = 5 and \( pK_b=4.745 \).
2. \( pOH=pK_b+\log([salt]/[base]) \) gives \( \log[salt]=0.255 \).
3. \( [salt]=10^{0.255}=1.8 \) mol per litre.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) 1 : 10
(b) 10 : 1
(c) 5 : 4
(d) 4 : 5
Show Answer & Explanation
Answer: (b) 10 : 1
Explanation:
1. \( pH=pK_a+\log([salt]/[acid]) \): \( 5=4+\log(\text{ratio}) \).
2. The ratio is 10 : 1.
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
(a) \( 2.63\times10^{-5} \)
(b) \( 3.0\times10^{-5} \)
(c) 4.68
(d) none of these
Show Answer & Explanation
Answer: (a) \( 2.63\times10^{-5} \)
Explanation:
1. \( [H^+]=10^{-4.58}=10^{0.42}\times10^{-5}=2.63\times10^{-5} \) mol \( L^{-1} \).
Teacher's Note:
1. Write the relevant expression (\( K_{sp} \), \( K_h \), pH or the Henderson equation) first, then substitute the data carefully.
2. Check that the answer is sensible: salts of strong bases are basic, salts of strong acids are acidic.
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