JEE Chemistry Ionic Equilibrium MCQs Set 05

Download JEE MCQs for JEE Chemistry: Ionic Equilibrium

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Chapter-wise Objective Questions: Ionic Equilibrium

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Multiple Choice Questions

Question 1: The solubility of \( \ce{BaSO4} \) in water is \( 2.33\times10^{-3} \) g/L (molar mass 233). Its solubility product is
(a) \( 1\times10^{-5} \)
(b) \( 1\times10^{-10} \)
(c) \( 1\times10^{-15} \)
(d) \( 1\times10^{-20} \)
Show Answer & Explanation

Answer: (b) \( 1\times10^{-10} \)

Explanation:
1. \( s=\frac{2.33\times10^{-3}}{233}=10^{-5} \) M.
2. \( K_{sp}=s^2=10^{-10} \).

Teacher's Note:
1. Divide by molar mass first.

Question 2: What is the \( \ce{H+} \) concentration when 4 g of NaOH is dissolved in 1000 mL of water?
(a) \( 10^{-1} \)
(b) \( 10^{-13} \)
(c) \( 10^{-4} \)
(d) \( 10^{-10} \)
Show Answer & Explanation

Answer: (b) \( 10^{-13} \)

Explanation:
1. \( [\ce{OH-}]=\frac{4}{40}=0.1 \) M.
2. \( [\ce{H+}]=\frac{10^{-14}}{0.1}=10^{-13} \) M.

Teacher's Note:
1. Use Kw for the hydrogen ion concentration.

Question 3: When 10 mL of 0.1 M acetic acid (\( pK_a=5.0 \)) is titrated against 10 mL of 0.1 M ammonia (\( pK_b=5.0 \)), the pH at the equivalence point is
(a) 5.0
(b) 6.0
(c) 7.0
(d) 9.0
Show Answer & Explanation

Answer: (c) 7.0

Explanation:
1. The salt formed is ammonium acetate with \( pK_a=pK_b \).
2. \( pH=7+\tfrac12(pK_a-pK_b)=7 \).

Teacher's Note:
1. Equal acid and base strengths give pH 7.

Question 4: The solubility product of \( \ce{BaSO4} \) at 25 C is \( 1.0\times10^{-9} \). The concentration of \( \ce{H2SO4} \) needed to precipitate \( \ce{BaSO4} \) from 0.01 M \( \ce{Ba^2+} \) is
(a) \( 10^{-9} \)
(b) \( 10^{-8} \)
(c) \( 10^{-7} \)
(d) \( 10^{-6} \)
Show Answer & Explanation

Answer: (c) \( 10^{-7} \)

Explanation:
1. Precipitation starts when \( [\ce{SO4^2-}]=\frac{10^{-9}}{0.01}=10^{-7} \) M.

Teacher's Note:
1. Ionic product must reach Ksp.

Question 5: On passing \( \ce{H2S} \) through a solution of \( \ce{Cu^2+} \) and \( \ce{Zn^2+} \) in acid, CuS is precipitated first because
(a) \( K_{sp} \) of CuS equals the ionic product of ZnS
(b) \( K_{sp} \) of CuS equals the \( K_{sp} \) of ZnS
(c) \( K_{sp} \) of CuS is lower than the \( K_{sp} \) of ZnS
(d) \( K_{sp} \) of CuS is greater than the \( K_{sp} \) of ZnS
Show Answer & Explanation

Answer: (c) \( K_{sp} \) of CuS is lower than the \( K_{sp} \) of ZnS

Explanation:
1. The ionic product reaches the very small Ksp of CuS at a much lower sulphide concentration than that of ZnS.

Teacher's Note:
1. Lower Ksp precipitates first.

Question 6: The solubility of \( \ce{MX2} \) type electrolytes is \( 0.5\times10^{-4} \) mol/L. Their \( K_{sp} \) is
(a) \( 5\times10^{-12} \)
(b) \( 25\times10^{-10} \)
(c) \( 1\times10^{-13} \)
(d) \( 5\times10^{-13} \)
Show Answer & Explanation

Answer: (d) \( 5\times10^{-13} \)

Explanation:
1. \( K_{sp}=4s^3=4(0.5\times10^{-4})^3=4\times1.25\times10^{-13}=5\times10^{-13} \).

Teacher's Note:
1. MX2 gives one M and two X ions.

Question 7: The solubility product of AgCl is \( 1\times10^{-10} \). The solubility of AgCl is
(a) \( 1\times10^{-10} \)
(b) \( 1\times10^{-7} \)
(c) \( 1\times10^{-5} \)
(d) \( 1\times10^{-8} \)
Show Answer & Explanation

Answer: (c) \( 1\times10^{-5} \)

Explanation:
1. \( s=\sqrt{K_{sp}}=\sqrt{10^{-10}}=10^{-5} \) mol/L.

Teacher's Note:
1. Square root for a 1:1 salt.

Question 8: Why is \( \ce{NH4OH} \) added to the solution before passing \( \ce{H2S} \) to precipitate zinc as ZnS?
(a) To convert Zn into \( \ce{Zn^2+} \)
(b) To reduce zinc
(c) To decrease the dissociation of \( \ce{H2S} \)
(d) To increase the dissociation of \( \ce{H2S} \)
Show Answer & Explanation

Answer: (d) To increase the dissociation of \( \ce{H2S} \)

Explanation:
1. \( \ce{OH-} \) removes \( \ce{H+} \) from \( \ce{H2S} \), shifting its ionisation forward and raising \( [\ce{S^2-}] \) enough to precipitate ZnS.

Teacher's Note:
1. A basic medium favours sulphide formation.

Question 9: Three reactions involve \( \ce{H2PO4-} \): (i) \( \ce{H3PO4 + H2O -> H3O+ + H2PO4-} \) (ii) \( \ce{H2PO4- + H2O -> HPO4^2- + H3O+} \) (iii) \( \ce{H2PO4- + OH- -> H3PO4 + O^2-} \). In which does \( \ce{H2PO4-} \) act as an acid?
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
(d) (iii) only
Show Answer & Explanation

Answer: (b) (ii) only

Explanation:
1. An acid donates a proton. In (ii) \( \ce{H2PO4-} \) loses a proton to water.
2. In (i) it is the product, and in (iii) it accepts a proton.

Teacher's Note:
1. Look at the side of the reaction where it appears and what it does.

Question 10: The \( K_{sp} \) of \( \ce{Cr(OH)3} \) is \( 1.6\times10^{-30} \). Its molar solubility is
(a) \( \sqrt{1.6\times10^{-30}} \)
(b) \( \sqrt[4]{1.6\times10^{-30}} \)
(c) \( \sqrt[4]{1.6\times10^{-30}/27} \)
(d) \( 1.6\times10^{-30}/27 \)
Show Answer & Explanation

Answer: (c) \( \sqrt[4]{1.6\times10^{-30}/27} \)

Explanation:
1. \( K_{sp}=s(3s)^3=27s^4 \).
2. \( s=\left(\frac{K_{sp}}{27}\right)^{1/4} \).

Teacher's Note:
1. Hydroxide is 3s, so the cube gives 27.

Question 11: How many litres of water must be added to 1 litre of aqueous HCl of pH 1 to obtain pH 2?
(a) 0.1 L
(b) 0.9 L
(c) 2.0 L
(d) 9.0 L
Show Answer & Explanation

Answer: (d) 9.0 L

Explanation:
1. A tenfold dilution is needed, so the final volume is 10 L.
2. Water added \( =10-1=9 \) L.

Teacher's Note:
1. Subtract the original volume from the final volume.

Question 12: What volume of 0.1 M \( \ce{H2SO4} \) is needed to neutralise 40 mL of 0.2 M NaOH completely?
(a) 10 mL
(b) 40 mL
(c) 20 mL
(d) 80 mL
Show Answer & Explanation

Answer: (b) 40 mL

Explanation:
1. NaOH \( =8 \) mmol; 0.1 M \( \ce{H2SO4} \) gives 0.2 mmol \( \ce{H+} \) per mL.
2. Volume \( =8/0.2=40 \) mL.

Teacher's Note:
1. Sulphuric acid is dibasic.

Question 13: The aqueous solution of which salt has the lowest pH?
(a) NaClO
(b) \( \ce{NaClO2} \)
(c) \( \ce{NaClO3} \)
(d) \( \ce{NaClO4} \)
Show Answer & Explanation

Answer: (d) \( \ce{NaClO4} \)

Explanation:
1. \( \ce{HClO4} \) is a strong acid, so \( \ce{ClO4-} \) does not hydrolyse and the solution is neutral.
2. The other anions come from weak acids and give basic solutions.

Teacher's Note:
1. Strongest parent acid: least basic salt.

Question 14: For an indicator HIn, the pH is related to the ratio of its acid and base forms by
(a) \( \log\frac{[\ce{HIn}]}{[\ce{In-}]}=pK_{In}-pH \)
(b) \( \log\frac{[\ce{HIn}]}{[\ce{In-}]}=pH-pK_{In} \)
(c) \( \log\frac{[\ce{In-}]}{[\ce{HIn}]}=pH+pK_{In} \)
(d) \( \log\frac{[\ce{In-}]}{[\ce{HIn}]}=pK_{In}-pH \)
Show Answer & Explanation

Answer: (a) \( \log\frac{[\ce{HIn}]}{[\ce{In-}]}=pK_{In}-pH \)

Explanation:
1. \( pH=pK_{In}+\log\frac{[\ce{In-}]}{[\ce{HIn}]} \).
2. Inverting the log ratio gives \( \log\frac{[\ce{HIn}]}{[\ce{In-}]}=pK_{In}-pH \).

Teacher's Note:
1. Henderson equation for an indicator.

Question 15: The correct order of acid strength is
(a) \( \ce{HClO4 \lt HClO3 \lt HClO2 \lt HClO} \)
(b) \( \ce{HClO2 \lt HClO3 \lt HClO4 \lt HClO} \)
(c) \( \ce{HClO4 \lt HClO \lt HClO2 \lt HClO3} \)
(d) \( \ce{HClO \lt HClO2 \lt HClO3 \lt HClO4} \)
Show Answer & Explanation

Answer: (d) \( \ce{HClO \lt HClO2 \lt HClO3 \lt HClO4} \)

Explanation:
1. Acid strength rises with the number of oxygen atoms on chlorine.
2. So \( \ce{HClO \lt HClO2 \lt HClO3 \lt HClO4} \).

Teacher's Note:
1. More oxygen atoms stabilise the anion.

Question 16: 10 mL of concentrated \( \ce{H2SO4} \) (18 M) is diluted to 1 litre. The normality of the diluted acid is
(a) 0.18 N
(b) 0.09 N
(c) 0.36 N
(d) 18 N
Show Answer & Explanation

Answer: (c) 0.36 N

Explanation:
1. Molarity after dilution \( =\frac{18\times10}{1000}=0.18 \) M.
2. Normality \( =0.18\times2=0.36 \) N.

Teacher's Note:
1. Normality is molarity times basicity.

Question 17: When HF is dissolved in formic acid, the equilibrium is \( \ce{HF + HCOOH \lt => F- + HCOOH2+} \). The true acid / conjugate base and base / conjugate acid pairs are
(a) (HF, HCOOH) and (\( \ce{HCOOH2+} \), \( \ce{F-} \))
(b) (HF, \( \ce{HCOOH2+} \)) and (HCOOH, \( \ce{F-} \))
(c) (\( \ce{HCOOH2+} \), HF) and (\( \ce{F-} \), HCOOH)
(d) (HF, \( \ce{F-} \)) and (\( \ce{HCOOH2+} \), HCOOH)
Show Answer & Explanation

Answer: (d) (HF, \( \ce{F-} \)) and (\( \ce{HCOOH2+} \), HCOOH)

Explanation:
1. HF donates a proton and becomes \( \ce{F-} \): acid / conjugate base.
2. HCOOH accepts a proton and becomes \( \ce{HCOOH2+} \): so the pair is (\( \ce{HCOOH2+} \), HCOOH) as conjugate acid / base.

Teacher's Note:
1. A conjugate pair differs by one proton.

Question 18: How many grams of a dibasic acid (molar mass 200) must be present in 100 mL of solution to make it N/10?
(a) 1 g
(b) 2 g
(c) 5 g
(d) 10 g
Show Answer & Explanation

Answer: (a) 1 g

Explanation:
1. Equivalent mass \( =200/2=100 \).
2. Equivalents needed \( =0.1\times0.1=0.01 \), so mass \( =0.01\times100=1 \) g.

Teacher's Note:
1. Mass = normality x volume (L) x equivalent mass.

Question 19: A solution contains equimolar \( \ce{NaHCO3} \) and NaOH. The species present are
(a) \( \ce{Na2CO3} \)
(b) \( \ce{NaHCO3} \) and NaOH
(c) NaOH
(d) \( \ce{NaHCO3} \) and \( \ce{Na2CO3} \)
Show Answer & Explanation

Answer: (a) \( \ce{Na2CO3} \)

Explanation:
1. \( \ce{NaHCO3 + NaOH -> Na2CO3 + H2O} \) goes to completion for equimolar amounts, leaving only \( \ce{Na2CO3} \).

Teacher's Note:
1. An acid salt and a base neutralise each other.

Question 20: An aqueous solution has \( [\ce{H+}]=10^{-4} \) M. It is diluted by mixing an equal volume of water. The concentration of \( \ce{OH-} \) in mol/dm3 will be
(a) \( 0.5\times10^{-10} \)
(b) \( 2\times10^{-10} \)
(c) \( 10^{-6} \)
(d) \( 10^{-8} \)
Show Answer & Explanation

Answer: (b) \( 2\times10^{-10} \)

Explanation:
1. After dilution \( [\ce{H+}]=5\times10^{-5} \) M.
2. \( [\ce{OH-}]=\frac{10^{-14}}{5\times10^{-5}}=2\times10^{-10} \) M.

Teacher's Note:
1. Find hydrogen ions first, then use Kw.

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