Practice MCQs for JEE Chemistry Ionic Equilibrium
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Multiple Choice Questions
(a) 0.0556 M
(b) 1.0 M
(c) 5.6 M
(d) 0.556 M
Show Answer & Explanation
Answer: (d) 0.556 M
Explanation:
1. \([OH^-]=10^{-2.5}=3.16\times10^{-3}\ \text{M}\).
2. \(C=\dfrac{[OH^-]^2}{K_b}=\dfrac{10^{-5}}{1.8\times10^{-5}}=0.556\ \text{M}\).
Teacher's Note:
1. Rearrange Kb = [OH-]^2 / C.
2. Check that the degree of ionisation is small.
(a) \(1\times10^{-5}\)
(b) \(1.11\times10^{-4}\)
(c) \(1.11\times10^{-5}\)
(d) \(1\times10^{-3}\)
Show Answer & Explanation
Answer: (b) \(1.11\times10^{-4}\)
Explanation:
1. \([H^+]=10^{-3}\ \text{M}\), so \(\alpha=10^{-3}/0.01=0.1\).
2. \(K_a=\dfrac{C\alpha^2}{1-\alpha}=\dfrac{0.01\times0.01}{0.9}=1.11\times10^{-4}\).
Teacher's Note:
1. Alpha is 10 percent here, so keep the (1 - alpha) term.
2. The simple formula C times alpha squared would give 1e-4.
(a) \([H^+]=2.0\times10^{-2}\ \text{M}\), \([A^-]=0.01\ \text{M}\)
(b) \([H^+]=1.0\times10^{-2}\ \text{M}\), \([A^-]=0\)
(c) \([H^+]=1.61\times10^{-2}\ \text{M}\), \([A^-]=6.13\times10^{-3}\ \text{M}\)
(d) \([H^+]=1.61\times10^{-2}\ \text{M}\), \([A^-]=1.0\times10^{-2}\ \text{M}\)
Show Answer & Explanation
Answer: (c) \([H^+]=1.61\times10^{-2}\ \text{M}\), \([A^-]=6.13\times10^{-3}\ \text{M}\)
Explanation:
1. Let \(x=[A^-]\). Then \([H^+]=0.01+x\) and \([HA]=0.01-x\).
2. \(\dfrac{x(0.01+x)}{0.01-x}=0.0255\) gives \(x^2+0.0355x-2.55\times10^{-4}=0\), so \(x=6.13\times10^{-3}\).
3. \([H^+]=0.01+0.00613=1.61\times10^{-2}\ \text{M}\).
Teacher's Note:
1. The acid is fairly strong, so the quadratic is needed.
2. The strong acid suppresses but does not stop the ionisation of the weak acid.
(a) 50%
(b) 0.1%
(c) 1%
(d) 10%
Show Answer & Explanation
Answer: (c) 1%
Explanation:
1. Neglecting water, \([OH^-]=10^{-6}\) and \([H^+]=10^{-8}\ \text{M}\).
2. Including water: \([H^+](10^{-6}+[H^+])=10^{-14}\) gives \([H^+]=9.90\times10^{-9}\ \text{M}\).
3. Error \(=(10^{-8}-9.90\times10^{-9})/9.90\times10^{-9}\approx1\%\).
Teacher's Note:
1. The error is small because 1e-6 is much larger than 1e-7.
2. At 1e-7 M the error would be huge.
(a) \(2\times10^{-3}\ \text{M},\ 3.6\times10^{-4}\ \text{M},\ 6.4\times10^{-4}\ \text{M}\)
(b) \(10^{-3}\ \text{M},\ 10^{-3}\ \text{M},\ 10^{-3}\ \text{M}\)
(c) \(10^{-3}\ \text{M},\ 6.4\times10^{-4}\ \text{M},\ 3.6\times10^{-4}\ \text{M}\)
(d) \(10^{-3}\ \text{M},\ 3.6\times10^{-4}\ \text{M},\ 6.4\times10^{-4}\ \text{M}\)
Show Answer & Explanation
Answer: (d) \(10^{-3}\ \text{M},\ 3.6\times10^{-4}\ \text{M},\ 6.4\times10^{-4}\ \text{M}\)
Explanation:
1. \([H^+]=\sqrt{1.8\times10^{-5}\times0.02+6.4\times10^{-5}\times0.01}=\sqrt{10^{-6}}=10^{-3}\ \text{M}\).
2. \([CH_3COO^-]=K_aC/[H^+]=3.6\times10^{-7}/10^{-3}=3.6\times10^{-4}\ \text{M}\).
3. \([C_7H_5O_2^-]=6.4\times10^{-7}/10^{-3}=6.4\times10^{-4}\ \text{M}\).
Teacher's Note:
1. The total anion concentration equals the hydrogen ion concentration.
2. 3.6e-4 plus 6.4e-4 equals 1e-3, a good check.
(a) \(\ce{HCOOK}\)
(b) \(\ce{NaH2PO4}\)
(c) \(\ce{CH3COONa}\)
(d) \(\ce{NaNO3}\)
Show Answer & Explanation
Answer: (b) \(\ce{NaH2PO4}\)
Explanation:
1. \(\ce{H2PO4-}\) is amphiprotic. Its acid constant \(K_{a2}\approx6\times10^{-8}\) is much larger than its base constant \(K_w/K_{a1}\approx10^{-12}\).
2. So its solution is acidic (pH about 4.7).
3. \(\ce{CH3COONa}\) and \(\ce{HCOOK}\) are basic (anion of a weak acid) and \(\ce{NaNO3}\) is neutral.
Teacher's Note:
1. The salt of a strong base and a weak acid is basic.
2. For an amphiprotic anion compare its Ka with its Kb.
(a) 4.48
(b) 2.48
(c) 7.00
(d) 9.52
Show Answer & Explanation
Answer: (a) 4.48
Explanation:
1. \(K_h=K_w/K_b=5.56\times10^{-10}\).
2. \([H^+]=\sqrt{K_hC}=\sqrt{5.56\times10^{-10}\times2}=3.33\times10^{-5}\ \text{M}\).
3. \(\text{pH}=4.48\).
Teacher's Note:
1. Ammonium chloride gives an acidic solution.
2. Use Kh = Kw/Kb.
(a) \(6.25\times10^{-6}\)
(b) \(6.25\times10^{-10}\)
(c) \(1.6\times10^{-9}\)
(d) \(1.6\times10^{-5}\)
Show Answer & Explanation
Answer: (b) \(6.25\times10^{-10}\)
Explanation:
1. \([H^+]=10^{-2.699}=2\times10^{-3}\ \text{M}\).
2. \(K_h=\dfrac{[H^+]^2}{C}=\dfrac{4\times10^{-6}}{0.25}=1.6\times10^{-5}\).
3. \(K_b=K_w/K_h=6.25\times10^{-10}\).
Teacher's Note:
1. Hydrolysis constant of the cation is Kw over Kb.
2. The cation of a weak base is the acid.
(a) 0.56% and pH 9.2
(b) 0.056% and pH 5
(c) 5.6% and pH 7
(d) 0.56% and pH 7
Show Answer & Explanation
Answer: (d) 0.56% and pH 7
Explanation:
1. \(K_h=\dfrac{K_w}{K_aK_b}=3.09\times10^{-5}\), and \(h=\sqrt{K_h}=5.56\times10^{-3}\) (independent of concentration).
2. \(h=0.56\%\).
3. \(\text{pH}=7+\tfrac12(\text{p}K_a-\text{p}K_b)=7\) because \(K_a=K_b\).
Teacher's Note:
1. For a salt of weak acid and weak base, h does not depend on concentration.
2. If Ka equals Kb the solution is neutral.
(a) 1.667%
(b) 0.167%
(c) 16.7%
(d) 5.0%
Show Answer & Explanation
Answer: (a) 1.667%
Explanation:
1. \(K_h=K_w/K_a=1.667\times10^{-5}\).
2. \(h=\sqrt{K_h/C}=\sqrt{2.78\times10^{-4}}=0.01667\).
3. Percent hydrolysis \(=1.667\%\).
Teacher's Note:
1. Convert the fraction to percent at the end.
2. Cyanide hydrolyses to give a basic solution.
(a) 0.025%
(b) 0.25%
(c) 2.5%
(d) 5%
Show Answer & Explanation
Answer: (b) 0.25%
Explanation:
1. \(\ce{CrO4^{2-} + H2O <=> HCrO4- + OH-}\) with \(K_h=K_w/K_2=3.23\times10^{-8}\).
2. \(h=\sqrt{K_h/C}=\sqrt{3.23\times10^{-8}/0.005}=2.54\times10^{-3}\).
3. The extent of hydrolysis is about 0.25%.
Teacher's Note:
1. The relevant acid constant is K2, since chromate is the second conjugate base.
2. Keep the answer as a percentage.
(a) 1.6%
(b) 0.4%
(c) 40%
(d) 4.0%
Show Answer & Explanation
Answer: (d) 4.0%
Explanation:
1. \(K_h=1.61\times10^{-5}\).
2. \(h=\sqrt{K_h/C}=\sqrt{1.61\times10^{-3}}=0.040\).
3. So hydrolysis is 4.0%.
Teacher's Note:
1. A tenfold lower concentration than the previous problem triples the percentage.
2. Check h is small enough for the approximation.
(a) \(K_h=10^{-2},\ K_b=10^{-12}\)
(b) \(K_h=10^{-4},\ K_b=10^{-10}\)
(c) \(K_h=10^{-6},\ K_b=10^{-8}\)
(d) \(K_h=10^{-8},\ K_b=10^{-6}\)
Show Answer & Explanation
Answer: (c) \(K_h=10^{-6},\ K_b=10^{-8}\)
Explanation:
1. \([H^+]=10^{-4}\ \text{M}\), so \(K_h=\dfrac{[H^+]^2}{C}=\dfrac{10^{-8}}{0.01}=10^{-6}\).
2. \(K_b\) of the conjugate base \(=K_w/K_h=10^{-14}/10^{-6}=10^{-8}\).
Teacher's Note:
1. The two constants multiply to Kw.
2. Do not mix up which constant belongs to the cation.
(a) 10.04
(b) 10.42
(c) 9.50
(d) 11.00
Show Answer & Explanation
Answer: (b) 10.42
Explanation:
1. \(K_h=K_w/K_a=9.52\times10^{-5}\), which is not small compared with the concentration.
2. Solve \(\dfrac{x^2}{10^{-3}-x}=9.52\times10^{-5}\), giving \([OH^-]=x=2.65\times10^{-4}\ \text{M}\).
3. \(\text{pOH}=3.58\), so \(\text{pH}=10.42\).
Teacher's Note:
1. Use the quadratic when h is about 25 percent.
2. The square-root shortcut would give a noticeably larger OH concentration.
(a) 8.35
(b) 10.35
(c) 7.00
(d) 6.35
Show Answer & Explanation
Answer: (a) 8.35
Explanation:
1. For an amphiprotic anion, \(\text{pH}=\tfrac12(\text{p}K_1+\text{p}K_2)\).
2. \(\text{p}K_1=6.35\) and \(\text{p}K_2=10.35\).
3. \(\text{pH}=8.35\).
Teacher's Note:
1. The pH of an amphiprotic salt is independent of its concentration.
2. Average the two pK values.
(a) 2.94
(b) 5.44
(c) 8.38
(d) 4.19
Show Answer & Explanation
Answer: (d) 4.19
Explanation:
1. The hydrogen phthalate ion is amphiprotic.
2. \(\text{pH}=\tfrac12(2.94+5.44)=4.19\).
Teacher's Note:
1. Use the amphiprotic formula.
2. The result lies between the two pK values.
(a) \(2.6\times10^{-9}\ \text{M}\)
(b) \(5.13\times10^{-6}\ \text{M}\)
(c) \(1\times10^{-7}\ \text{M}\)
(d) \(1.9\times10^{-5}\ \text{M}\)
Show Answer & Explanation
Answer: (b) \(5.13\times10^{-6}\ \text{M}\)
Explanation:
1. Equal volumes are mixed, so the salt concentration is \(0.05\ \text{M}\).
2. \([OH^-]=\sqrt{\dfrac{K_w}{K_a}C}=\sqrt{\dfrac{10^{-14}}{1.9\times10^{-5}}\times0.05}\).
3. \([OH^-]=5.13\times10^{-6}\ \text{M}\).
Teacher's Note:
1. At equivalence the volume has doubled.
2. The equivalence point of a weak acid with a strong base is basic.
(a) pH \(=8\) and \(K_b=1\times10^{-5}\)
(b) pH \(=6\) and \(K_b=1\times10^{-5}\)
(c) pH \(=6\) and \(K_b=1\times10^{-9}\)
(d) pH \(=3\) and \(K_b=1\times10^{-14}\)
Show Answer & Explanation
Answer: (b) pH \(=6\) and \(K_b=1\times10^{-5}\)
Explanation:
1. \([H^+]=\sqrt{K_hC}=\sqrt{10^{-9}\times10^{-3}}=10^{-6}\ \text{M}\), so \(\text{pH}=6\).
2. \(K_b=K_w/K_h=10^{-14}/10^{-9}=10^{-5}\).
Teacher's Note:
1. The hydrated zinc ion is a weak acid.
2. Kb of the conjugate base is Kw over Kh.
(a) 9.12
(b) 8.82
(c) 4.88
(d) 9.48
Show Answer & Explanation
Answer: (a) 9.12
Explanation:
1. Base \(=40\) mmol and salt \(=60\) mmol.
2. \(\text{pOH}=\text{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}=4.70+\log1.5=4.88\).
3. \(\text{pH}=14-4.88=9.12\).
Teacher's Note:
1. A mixture of a weak base and its salt is a basic buffer.
2. Use pOH = pKb + log(salt/base).
(a) 4.70
(b) 7.00
(c) 9.30
(d) 12.48
Show Answer & Explanation
Answer: (d) 12.48
Explanation:
1. \(\ce{NaOH}=30\) mmol and \(\ce{CH3COOH}=15\) mmol.
2. The acid is fully neutralised and \(15\) mmol of \(\ce{NaOH}\) is left in 500 mL, so \([OH^-]=0.03\ \text{M}\).
3. \(\text{pOH}=1.52\), so \(\text{pH}=12.48\). The acetate formed is a very weak base and does not matter.
Teacher's Note:
1. Check which reagent is in excess before choosing a formula.
2. Excess strong base controls the pH.
(a) 5.06
(b) 9.00
(c) 8.94
(d) 4.70
Show Answer & Explanation
Answer: (a) 5.06
Explanation:
1. \(\ce{NH4OH}=120\) mmol and \(\ce{HCl}=120\) mmol, so exactly neutralised to \(\ce{NH4Cl}\) of concentration \(120/800=0.15\ \text{M}\).
2. \(K_h=K_w/K_b=5\times10^{-10}\), so \([H^+]=\sqrt{5\times10^{-10}\times0.15}=8.66\times10^{-6}\ \text{M}\).
3. \(\text{pH}=5.06\).
Teacher's Note:
1. Exact neutralisation of a weak base leaves an acidic salt solution.
2. Use the salt hydrolysis formula, not the buffer formula.
(a) \(3.6\times10^{-5}\ \text{M}\)
(b) \(9\times10^{-6}\ \text{M}\)
(c) \(1.8\times10^{-5}\ \text{M}\)
(d) \(9.5\times10^{-4}\ \text{M}\)
Show Answer & Explanation
Answer: (b) \(9\times10^{-6}\ \text{M}\)
Explanation:
1. \([OH^-]=K_b\dfrac{[NH_3]}{[NH_4^+]}=1.8\times10^{-5}\times\dfrac{0.050}{0.100}\).
2. \([OH^-]=9.0\times10^{-6}\ \text{M}\).
Teacher's Note:
1. Common ion effect suppresses the ionisation of ammonia.
2. The ratio base to salt appears directly.
(a) 9.26
(b) 2.87
(c) 4.74
(d) 7.00
Show Answer & Explanation
Answer: (c) 4.74
Explanation:
1. \(\ce{CH3COOH}=10\) mmol and \(\ce{NaOH}=5\) mmol.
2. After reaction there are 5 mmol of acid and 5 mmol of acetate.
3. \(\text{pH}=\text{p}K_a=4.74\).
Teacher's Note:
1. Half neutralisation of a weak acid gives pH equal to pKa.
2. Equal amounts of acid and salt make the log term zero.
(a) 4.76
(b) 11.24
(c) 9.24
(d) 8.24
Show Answer & Explanation
Answer: (c) 9.24
Explanation:
1. \([OH^-]=\sqrt{K_bC}=\sqrt{1.5\times10^{-9}\times0.2}=1.73\times10^{-5}\ \text{M}\).
2. \(\text{pOH}=4.76\), so \(\text{pH}=9.24\).
Teacher's Note:
1. Pyridine is a very weak base.
2. Find pOH first.
(a) 9.30
(b) 5.70
(c) 4.70
(d) 9.24
Show Answer & Explanation
Answer: (c) 4.70
Explanation:
1. \(\text{p}K_b=8.82\).
2. \(\text{pOH}=\text{p}K_b+\log\dfrac{[\text{salt}]}{[\text{base}]}=8.82+\log3=9.30\).
3. \(\text{pH}=14-9.30=4.70\).
Teacher's Note:
1. Adding the conjugate acid lowers the pH.
2. This is a basic buffer in which the salt predominates.
(a) 9.26
(b) 4.74
(c) 8.78
(d) 5.22
Show Answer & Explanation
Answer: (c) 8.78
Explanation:
1. \(\ce{HCl}=15\) mmol and \(\ce{NH3}=20\) mmol, so \(\ce{NH3}=5\) mmol and \(\ce{NH4+}=15\) mmol remain.
2. \(\text{pOH}=4.74+\log(15/5)=5.22\).
3. \(\text{pH}=14-5.22=8.78\).
Teacher's Note:
1. Acid is limiting, so ammonia is left over with its salt.
2. This is a basic buffer.
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