JEE Chemistry Ionic Equilibrium MCQs Set 10

Answer: (d) \(\ce{CH3OH}\) is the acid and \(\ce{H-}\) is the base

Explanation:
1. \(\ce{CH3OH}\) loses a proton and becomes \(\ce{CH3O-}\), so it is the acid.
2. \(\ce{H-}\) accepts the proton to give \(\ce{H2}\), so it is the base.
3. The hydride ion is a very strong base.

Teacher's Note:
1. Hydrogen gas is evolved when the hydride picks up the proton.
2. Alcohols can behave as acids toward very strong bases.

Answer: (a) \(\ce{NH3}\), \(\ce{HF}\) and \(\ce{N2H6^{2+}}\)

Explanation:
1. The conjugate acid is formed by adding one proton to the base.
2. \(\ce{NH2^- + H^+ -> NH3}\), \(\ce{F^- + H^+ -> HF}\), \(\ce{N2H5^+ + H^+ -> N2H6^{2+}}\).
3. Adding one proton raises the charge by one unit.

Teacher's Note:
1. Add one \(\ce{H+}\) to the base and never change anything else.
2. The conjugate base is formed by removing one proton instead.

Answer: (c) \(\ce{H2PO2-}\), \(\ce{SO4^{2-}}\), \(\ce{HCO3-}\) and \(\ce{ClO4-}\)

Explanation:
1. The conjugate base is the acid minus one proton.
2. \(\ce{H3PO2 -> H2PO2- + H+}\), \(\ce{HSO4- -> SO4^{2-} + H+}\), \(\ce{H2CO3 -> HCO3- + H+}\), \(\ce{HClO4 -> ClO4- + H+}\).
3. Hypophosphorous acid is monobasic, so only one proton is lost.

Teacher's Note:
1. Remove only one proton and reduce the charge by one unit.
2. Remember that hypophosphorous acid is monobasic.

Answer: (a) \(K_w=4\times10^{-14}\), pH \(=6.7\)

Explanation:
1. Degree of ionisation \(\alpha=3.6\times10^{-7}/100=3.6\times10^{-9}\).
2. \([H^+]=55.5\times3.6\times10^{-9}=2\times10^{-7}\ \text{M}\), so \(K_w=[H^+]^2=4\times10^{-14}\).
3. \(\text{pH}=-\log(2\times10^{-7})=7-0.30=6.7\).

Teacher's Note:
1. Pure water is neutral at every temperature, so [H+] equals [OH-].
2. Kw above 1e-14 means a temperature above 25 degrees C.

Answer: (b) 13.09

Explanation:
1. \(\ce{OH-}\) from base \(=20\times0.2\times2=8\) mmol and \(\ce{H+}\) from acid \(=25\times0.1=2.5\) mmol.
2. Excess \(\ce{OH-}=5.5\) mmol in 45 mL, so \([OH^-]=0.1222\ \text{M}\).
3. \(\text{pOH}=0.913\), so \(\text{pH}=13.09\).

Teacher's Note:
1. Each formula unit of calcium hydroxide supplies two hydroxide ions.
2. Find the excess of the reagent first, then divide by the total volume.

Answer: (b) 1.30

Explanation:
1. \(\ce{H+}=10\times0.1\times2=2\) mmol and \(\ce{OH-}=1\) mmol.
2. Excess \(\ce{H+}=1\) mmol in 20 mL gives \([H^+]=0.05\ \text{M}\).
3. \(\text{pH}=-\log0.05=1.30\).

Teacher's Note:
1. Treat sulphuric acid as fully dissociated in its first step and in this strong dilute case also the second.
2. Divide the excess by the total volume.

Answer: (d) 4.52

Explanation:
1. Initial moles of \(\ce{H+}=10^{-4}\) and the final volume is \(1+7/3=10/3\) L.
2. \([H^+]=10^{-4}\times3/10=3\times10^{-5}\ \text{M}\).
3. \(\text{pH}=5-\log3=5-0.48=4.52\).

Teacher's Note:
1. Dilution lowers concentration, so the pH goes up.
2. Use the number of moles because it does not change on dilution.

Answer: (d) 0.62

Explanation:
1. \(K_a=\dfrac{C\alpha^2}{1-\alpha}\) gives \(\dfrac{\alpha^2}{1-\alpha}=1\) because \(C=K_a\).
2. \(\alpha^2+\alpha-1=0\), so \(\alpha=\dfrac{-1+\sqrt5}{2}\).
3. \(\alpha=0.618\approx0.62\).

Teacher's Note:
1. The approximation alpha much less than 1 fails here, so solve the quadratic.
2. Whenever C equals Ka the quadratic gives the golden ratio 0.618.

Answer: (a) \(K_a=9\times10^{-9},\ K_b=9\times10^{-7}\)

Explanation:
1. Acid: \([H^+]=10^{-4.52}=3\times10^{-5}\ \text{M}\), so \(K_a=\dfrac{(3\times10^{-5})^2}{0.10}=9\times10^{-9}\).
2. Base: \(\text{pOH}=14-10.48=3.52\), \([OH^-]=3\times10^{-4}\ \text{M}\).
3. \(K_b=\dfrac{(3\times10^{-4})^2}{0.10}=9\times10^{-7}\).

Teacher's Note:
1. For a weak electrolyte use K = [ion]^2 / C.
2. The base needs pOH, not pH.

Answer: (b) 10

Explanation:
1. For a weak acid \(\alpha=\sqrt{K_a/C}\).
2. \(\alpha_2/\alpha_1=\sqrt{C_1/C_2}=\sqrt{100}\).
3. The ratio is about 10.

Teacher's Note:
1. Dilution increases dissociation, so the ratio exceeds one.
2. Ostwald dilution law: alpha is proportional to one over the square root of C.

Answer: (d) 170.4

Explanation:
1. At equal concentration \(\alpha\propto\sqrt{K_a}\).
2. Ratio \(=\sqrt{1.8\times10^{-5}/6.2\times10^{-10}}=\sqrt{29032}\).
3. The ratio is 170.4.

Teacher's Note:
1. Do not forget the square root when comparing degrees of dissociation.
2. The ratio of Ka values itself is 29032, which is a common wrong answer.

Answer: (b) \(K_a=10^{-8},\ K_b=10^{-6}\)

Explanation:
1. Acid: \([H^+]=10^{-4.5}\), \(K_a=\dfrac{10^{-9}}{0.1}=10^{-8}\).
2. Base: \(\text{pOH}=3.5\), \([OH^-]=10^{-3.5}\).
3. \(K_b=\dfrac{10^{-7}}{0.1}=10^{-6}\).

Teacher's Note:
1. Square the ion concentration and divide by the molarity.
2. Work out pOH for the base.

Answer: (d) 1.00

Explanation:
1. HCl is a strong acid, so \([H^+]=0.1\ \text{M}\).
2. \(\text{pH}=-\log0.1=1\).

Teacher's Note:
1. Strong monoprotic acids are completely ionised.
2. The concentration of acid equals the concentration of hydrogen ions.

Answer: (c) 0.52

Explanation:
1. \(\ce{H+}\) from \(\ce{H2SO4}=50\times0.1\times2=10\) mmol and from \(\ce{HCl}=50\times0.4=20\) mmol.
2. Total 30 mmol in 100 mL, so \([H^+]=0.3\ \text{M}\).
3. \(\text{pH}=-\log0.3=0.52\).

Teacher's Note:
1. Add the millimoles of hydrogen ion from both acids.
2. Sulphuric acid supplies two protons per molecule.

Answer: (c) 2.87

Explanation:
1. \([H^+]=\sqrt{K_aC}=\sqrt{1.8\times10^{-5}\times0.1}=1.34\times10^{-3}\ \text{M}\).
2. \(\text{pH}=-\log(1.34\times10^{-3})=2.87\).

Teacher's Note:
1. Weak acid: use the square-root formula.
2. The pH of a weak acid lies between that of a strong acid of the same concentration and 7.

Answer: (a) 7.00

Explanation:
1. At this tiny concentration the hydroxide from water dominates.
2. \([OH^-]=10^{-10}+[H^+]\) gives \([OH^-]\approx1.0\times10^{-7}\ \text{M}\).
3. The solution is practically neutral, \(\text{pH}\approx7\).

Teacher's Note:
1. A base cannot give a pH below 7.
2. Water supplies 1e-7 M hydroxide by itself.

Answer: (c) 6.02

Explanation:
1. The acid is far more dilute than \(K_a\), so it is almost completely ionised: \(\alpha\approx1\).
2. \([H^+]\approx10^{-6}\ \text{M}\) (solving the full equation, including water, gives \(9.6\times10^{-7}\ \text{M}\)).
3. \(\text{pH}\approx6.02\).

Teacher's Note:
1. Square-root formula fails at very low concentration.
2. Check whether alpha is close to one before using it.

Answer: (d) 6.98

Explanation:
1. The acid is almost fully ionised but contributes only about \(10^{-8}\) M \(\ce{H+}\).
2. Water supplies \(10^{-7}\) M, so the total is \([H^+]\approx1.05\times10^{-7}\ \text{M}\).
3. \(\text{pH}\approx6.98\), almost identical to the same concentration of HCl.

Teacher's Note:
1. Acids always give pH less than 7 at 25 degrees C.
2. Water autoionisation dominates in very dilute solutions.

Answer: (c) 2.61

Explanation:
1. For a mixture of two weak acids \([H^+]=\sqrt{K_1C_1+K_2C_2}\).
2. \([H^+]=\sqrt{2\times10^{-6}+4\times10^{-6}}=\sqrt{6\times10^{-6}}=2.45\times10^{-3}\ \text{M}\).
3. \(\text{pH}=2.61\).

Teacher's Note:
1. Each acid adds its own Ka times C under the root.
2. Both acids repress each other slightly, which the approximation ignores.

Answer: (d) 11.30

Explanation:
1. Diluted concentration \(=0.1/100=10^{-3}\) M.
2. \([OH^-]=2\times10^{-3}\ \text{M}\) since there are two hydroxides per unit.
3. \(\text{pOH}=2.70\), so \(\text{pH}=11.30\).

Teacher's Note:
1. Barium hydroxide gives two hydroxide ions.
2. Decimolar means 0.1 M.

Answer: (d) 9.00

Explanation:
1. \([OH^-]=10^{-3}/100=10^{-5}\ \text{M}\).
2. \(\text{pOH}=5\), so \(\text{pH}=9\).

Teacher's Note:
1. Divide the moles by the volume in litres first.
2. Strong base is fully ionised.

Answer: (c) 3.00

Explanation:
1. \([H^+]\) values are \(10^{-4}\) and \(1.9\times10^{-3}\ \text{M}\).
2. For equal volumes the average is \((0.1+1.9)\times10^{-3}/2=10^{-3}\ \text{M}\).
3. \(\text{pH}=3\).

Teacher's Note:
1. Average the concentrations, never the pH values.
2. For a strong monoprotic acid normality equals molarity.

Answer: (c) 6.80

Explanation:
1. In neutral water \([H^+]=[OH^-]=\sqrt{K_w}=1.6\times10^{-7}\ \text{M}\).
2. \(\text{pH}=7-\log1.6=7-0.204=6.80\).

Teacher's Note:
1. Neutral pH is 7 only at 25 degrees C.
2. Kw rises with temperature, so neutral pH falls.

Answer: (d) \(6.022\times10^{7}\)

Explanation:
1. \([H^+]=10^{-13}\ \text{mol L}^{-1}=10^{-16}\ \text{mol mL}^{-1}\).
2. Number of ions \(=10^{-16}\times6.022\times10^{23}=6.022\times10^{7}\).

Teacher's Note:
1. Convert from per litre to per millilitre.
2. Multiply the moles by the Avogadro number.

Answer: (b) \(6\times10^{-8}\ \text{M}\ \text{(increase)}\)

Explanation:
1. At 298 K, \([H^+]=\sqrt{10^{-14}}=1.0\times10^{-7}\ \text{M}\).
2. At 310 K, \([H^+]=\sqrt{2.56\times10^{-14}}=1.6\times10^{-7}\ \text{M}\).
3. The increase is \(0.6\times10^{-7}=6\times10^{-8}\ \text{M}\).

Teacher's Note:
1. Water ionisation is endothermic, so [H+] increases with temperature.
2. Take the difference of the two square roots.

Answer: (d) \(2.31\times10^{-8}\ \text{M}\)

Explanation:
1. \([H^+]=10^{-6.95}=1.122\times10^{-7}\ \text{M}\).
2. Charge balance: \([H^+]=C+[OH^-]\) with \([OH^-]=10^{-14}/1.122\times10^{-7}=8.91\times10^{-8}\).
3. \(C=1.122\times10^{-7}-8.91\times10^{-8}=2.31\times10^{-8}\ \text{M}\).

Teacher's Note:
1. Do not ignore the hydrogen ions from water at this dilution.
2. The acid supplies only the difference between total H+ and OH-.