JEE Chemistry Ionic Equilibrium MCQs Set 03

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Multiple Choice Questions

Question 1: The molar concentration of chloride ion in the solution obtained by mixing 300 mL of 3.0 M NaCl and 200 mL of 4.0 M \( \ce{BaCl2} \) is
(a) 5.0 M
(b) 1.8 M
(c) 1.6 M
(d) none of these
Show Answer & Explanation

Answer: (a) 5.0 M

Explanation:
1. Chloride from NaCl \( =300\times3=900 \) mmol.
2. Chloride from \( \ce{BaCl2} \) \( =200\times4\times2=1600 \) mmol.
3. \( [\ce{Cl-}]=\frac{2500}{500}=5.0 \) M.

Teacher's Note:
1. Each formula unit of BaCl2 gives two chloride ions.

Question 2: A metal hydroxide \( \ce{M(OH)4} \) is 50% ionised. The pH of its 0.0025 M solution is
(a) 12
(b) 2
(c) 4
(d) 11.7
Show Answer & Explanation

Answer: (d) 11.7

Explanation:
1. \( [\ce{OH-}]=4\times0.0025\times0.5=5\times10^{-3} \) M.
2. pOH = 3 - 0.70 = 2.30, so pH = 11.7.

Teacher's Note:
1. Each formula unit gives four hydroxide ions.

Question 3: A \( 10^{-6} \) M HCl solution is diluted 100 times. Its pH is approximately
(a) 6.0
(b) 8.0
(c) 6.95
(d) 9.5
Show Answer & Explanation

Answer: (c) 6.95

Explanation:
1. The diluted acid is \( 10^{-8} \) M, so water's own ionisation cannot be ignored.
2. \( [\ce{H+}]=\frac{10^{-8}+\sqrt{10^{-16}+4\times10^{-14}}}{2}\approx1.05\times10^{-7} \) M.
3. pH is about 6.98, nearest to 6.95; the solution stays acidic.

Teacher's Note:
1. A very dilute acid can never give a pH above 7.

Question 4: The pH of a solution is increased from 3 to 6. Its \( \ce{H+} \) concentration will
(a) be reduced to half
(b) be doubled
(c) be reduced 1000 times
(d) increase 1000 times
Show Answer & Explanation

Answer: (c) be reduced 1000 times

Explanation:
1. \( [\ce{H+}] \) changes from \( 10^{-3} \) to \( 10^{-6} \) M.
2. It is reduced 1000 times.

Teacher's Note:
1. Each pH unit is a factor of 10.

Question 5: Which of the statements is/are correct? (a) The pH of \( 1.0\times10^{-8} \) M HCl is 8. (b) The conjugate base of \( \ce{H2PO4-} \) is \( \ce{HPO4^2-} \). (c) The autoprotolysis constant of water increases with temperature. (d) When a weak monoprotic acid is titrated with a strong base, the pH at half neutralisation is \( \tfrac12pK_a \).
(a) a only
(b) a, b
(c) a, b, d
(d) b, c
Show Answer & Explanation

Answer: (d) b, c

Explanation:
1. (a) False: an acid cannot give pH above 7; it is about 6.98.
2. (b) True.
3. (c) True: ionisation of water is endothermic, so Kw rises with T.
4. (d) False: at half neutralisation pH = pKa.

Teacher's Note:
1. Check every statement separately.

Question 6: An aqueous solution of strong acid of pH 5 is mixed with an equal volume of strong acid of pH 3. The pH of the mixture is
(a) 3.3
(b) 3.5
(c) 4.5
(d) 4.0
Show Answer & Explanation

Answer: (a) 3.3

Explanation:
1. Mean \( [\ce{H+}]=\frac{10^{-5}+10^{-3}}{2}\approx5.05\times10^{-4} \) M.
2. pH = 4 - log 5.05 = 3.3.

Teacher's Note:
1. Average the concentrations, not the pH values.

Question 7: \( \ce{H2X} \) is a dibasic acid that dissociates completely in water. The molarity of its aqueous solution of pH 1 is
(a) 0.1
(b) 0.05
(c) 0.2
(d) 0.5
Show Answer & Explanation

Answer: (b) 0.05

Explanation:
1. pH 1 means \( [\ce{H+}]=0.1 \) M.
2. Each molecule gives two \( \ce{H+} \), so \( C=0.05 \) M.

Teacher's Note:
1. Divide by the basicity.

Question 8: How many moles of HCl must be removed from 1 litre of aqueous HCl solution to change its pH from 2 to 3?
(a) 1
(b) 0.02
(c) 0.009
(d) 0.01
Show Answer & Explanation

Answer: (c) 0.009

Explanation:
1. Initial moles \( =10^{-2} \); final moles \( =10^{-3} \).
2. Removed \( =0.01-0.001=0.009 \) mol.

Teacher's Note:
1. Use moles in the same volume of 1 L.

Question 9: 8 g of NaOH and 4.9 g of \( \ce{H2SO4} \) are present in one litre of solution. Its pH is
(a) 1
(b) 13
(c) 12
(d) 2
Show Answer & Explanation

Answer: (b) 13

Explanation:
1. NaOH \( =8/40=0.2 \) mol; \( \ce{H2SO4}=4.9/98=0.05 \) mol, giving 0.1 mol \( \ce{H+} \).
2. Excess \( \ce{OH-}=0.1 \) M, so pOH = 1 and pH = 13.

Teacher's Note:
1. Neutralise first, then find the excess.

Question 10: The pH of a solution whose 100 mL contains 0.2 g NaOH is
(a) 10.699
(b) 11.699
(c) 12.699
(d) 13.699
Show Answer & Explanation

Answer: (c) 12.699

Explanation:
1. \( [\ce{OH-}]=\frac{0.2/40}{0.1}=0.05 \) M.
2. pOH = 1.301, so pH = 12.699.

Teacher's Note:
1. Convert grams to moles and then to molarity.

Question 11: One litre of solution contains 1 M HOCl (\( K_a=10^{-8} \)) and 1 M NaOH. The pH of the solution is
(a) 8
(b) 11
(c) 5
(d) 2
Show Answer & Explanation

Answer: (b) 11

Explanation:
1. They react completely to give 1 M NaOCl.
2. \( pH=7+\tfrac12(pK_a+\log C)=7+\tfrac12(8+0)=11 \).

Teacher's Note:
1. The final solution is a salt of a weak acid and strong base.

Question 12: What is the quantity of NaOH present in 250 mL of solution so that it gives pH = 13?
(a) \( 10^{-13} \) g
(b) \( 10^{-1} \) g
(c) 1.0 g
(d) 4.0 g
Show Answer & Explanation

Answer: (c) 1.0 g

Explanation:
1. pH 13 means \( [\ce{OH-}]=0.1 \) M.
2. Moles in 0.25 L \( =0.025 \); mass \( =0.025\times40=1.0 \) g.

Teacher's Note:
1. Moles equal molarity times volume in litres.

Question 13: To 50 mL of 0.05 M formic acid, what volume of 0.10 M sodium formate must be added to get a buffer of pH 4.0? (\( pK_a=3.8 \))
(a) 50 mL
(b) 4 mL
(c) 39.6 mL
(d) 100 mL
Show Answer & Explanation

Answer: (c) 39.6 mL

Explanation:
1. \( 4.0=3.8+\log\frac{[\text{salt}]}{[\text{acid}]} \), so ratio \( =10^{0.2}=1.585 \).
2. Acid \( =2.5 \) mmol, so salt \( =3.96 \) mmol.
3. Volume \( =3.96/0.10=39.6 \) mL.

Teacher's Note:
1. Use the Henderson equation with moles.

Question 14: \( pK_b \) of \( \ce{NH4OH} \) at a certain temperature is 4.74. The pH of a basic buffer with equimolar \( \ce{NH4OH} \) and \( \ce{NH4Cl} \) is
(a) 7.74
(b) 4.74
(c) 2.37
(d) 9.26
Show Answer & Explanation

Answer: (d) 9.26

Explanation:
1. For equal concentrations, \( pOH=pK_b=4.74 \).
2. pH = 14 - 4.74 = 9.26.

Teacher's Note:
1. Remember pH + pOH = 14.

Question 15: \( K_a \) of HCN is \( 5\times10^{-10} \) at 25 C. To maintain a pH of 9, what volume of 5 M KCN solution must be added to 10 mL of 2 M HCN?
(a) 4 mL
(b) 7.95 mL
(c) 2 mL
(d) 9.3 mL
Show Answer & Explanation

Answer: (c) 2 mL

Explanation:
1. \( pK_a=9.30 \); \( 9=9.30+\log\frac{[\text{salt}]}{[\text{acid}]} \), so salt/acid = 0.5.
2. Acid \( =20 \) mmol, so salt \( =10 \) mmol.
3. Volume \( =10/5=2 \) mL.

Teacher's Note:
1. Work with millimoles.

Question 16: The \( \ce{H+} \) concentration of 0.001 M acetic acid is \( 1.34\times10^{-4} \) g ion/L. What is the \( \ce{H+} \) concentration when 0.164 g of \( \ce{CH3COONa} \) is added to one litre of 0.001 M acetic acid?
(a) \( 9\times10^{-6} \)
(b) \( 18\times10^{-6} \)
(c) \( 4.5\times10^{-6} \)
(d) \( 5\times10^{-6} \)
Show Answer & Explanation

Answer: (a) \( 9\times10^{-6} \)

Explanation:
1. \( K_a=\frac{(1.34\times10^{-4})^2}{0.001}=1.8\times10^{-5} \).
2. Sodium acetate \( =0.164/82=0.002 \) M.
3. \( [\ce{H+}]=K_a\frac{0.001}{0.002}=9\times10^{-6} \).

Teacher's Note:
1. Common ion effect: the acetate suppresses ionisation.

Question 17: 50 mL of 2 N acetic acid is mixed with 10 mL of 1 N sodium acetate. The approximate pH is (\( K_a=10^{-5} \))
(a) 4
(b) 5
(c) 6
(d) 7
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Acid \( =100 \) meq, salt \( =10 \) meq.
2. \( pH=5+\log\frac{10}{100}=4 \).

Teacher's Note:
1. Use Henderson's equation with equivalents.

Question 18: On adding NaOH to acetic acid, 60% of the acid is neutralised. If \( pK_a \) is 4.7, the pH of the resulting solution is
(a) more than 4.7 but less than 5.0
(b) less than 4.7 but more than 4.0
(c) more than 5.0
(d) unchanged
Show Answer & Explanation

Answer: (a) more than 4.7 but less than 5.0

Explanation:
1. \( pH=4.7+\log\frac{60}{40}=4.7+0.18=4.88 \).
2. This lies between 4.7 and 5.0.

Teacher's Note:
1. The salt is 60 percent and the acid 40 percent.

Question 19: 500 mL of 0.2 M acetic acid is added to 500 mL of 0.30 M sodium acetate. If \( K_a=1.5\times10^{-5} \), the pH of the resulting solution is
(a) 5.0
(b) 9.0
(c) 3.0
(d) 4.0
Show Answer & Explanation

Answer: (a) 5.0

Explanation:
1. Acid \( =100 \) mmol, salt \( =150 \) mmol.
2. \( pK_a=4.824 \); \( pH=4.824+\log1.5=5.0 \).

Teacher's Note:
1. Equal volumes: the ratio of moles equals the ratio of the concentrations.

Question 20: Half of a formic acid solution is neutralised by KOH. If \( K_a \) of HCOOH is \( 2\times10^{-4} \), the pH of the solution is (log 2 = 0.3010)
(a) 3.6990
(b) 10.3010
(c) 3.85
(d) 4.3010
Show Answer & Explanation

Answer: (a) 3.6990

Explanation:
1. At half neutralisation [salt] = [acid], so pH = pKa.
2. \( pK_a=4-\log2=3.699 \).

Teacher's Note:
1. Half-neutralisation point: pH equals pKa.

Question 21: Calculate the pH of a buffer made by mixing 300 cc of 0.3 M \( \ce{NH3} \) and 500 cc of 0.5 M \( \ce{NH4Cl} \). \( K_b=1.8\times10^{-5} \).
(a) 8.1187
(b) 9.8117
(c) 8.8117
(d) none of these
Show Answer & Explanation

Answer: (c) 8.8117

Explanation:
1. \( \ce{NH3}=90 \) mmol, \( \ce{NH4Cl}=250 \) mmol.
2. \( pOH=4.745+\log\frac{250}{90}=5.188 \).
3. pH = 14 - 5.188 = 8.81.

Teacher's Note:
1. Basic buffer: use pOH = pKb + log(salt/base).

Question 22: What concentration of sodium propanoate must be added to one litre of 0.02 M propanoic acid (\( K_a=1.34\times10^{-5} \)) to get a buffer of pH 4.75?
(a) \( 4.52\times10^{-2} \) M
(b) \( 3.52\times10^{-2} \) M
(c) \( 2.52\times10^{-2} \) M
(d) \( 1.52\times10^{-2} \) M
Show Answer & Explanation

Answer: (d) \( 1.52\times10^{-2} \) M

Explanation:
1. \( pK_a=4.873 \); \( 4.75=4.873+\log\frac{S}{0.02} \).
2. \( \frac{S}{0.02}=10^{-0.123}=0.753 \), so \( S=1.5\times10^{-2} \) M.

Teacher's Note:
1. Rearrange the Henderson equation for the salt.

Question 23: 0.01 mole of HCl is dissolved in a buffer containing 0.02 mole of propanoic acid (\( K_a=1.34\times10^{-5} \)) and 0.0152 mole of its salt. The pH is [log 0.173 = -0.76]
(a) 3.11
(b) 4.11
(c) 5.11
(d) 6.11
Show Answer & Explanation

Answer: (b) 4.11

Explanation:
1. HCl converts 0.01 mol of salt to acid: salt \( =0.0052 \), acid \( =0.03 \).
2. \( pH=4.873+\log\frac{0.0052}{0.03}=4.873-0.76=4.11 \).

Teacher's Note:
1. Strong acid added to a buffer converts salt into acid.

Question 24: A buffer is made by dissolving 30 g of \( \ce{Na2CO3} \) in 500 mL of solution containing 150 mL of 1 M HCl (150 mmol). \( K_a \) of \( \ce{HCO3-} \) is \( 5.63\times10^{-11} \). [\( \log\frac{133}{150}=-0.05 \)] The pH is
(a) 8.197
(b) 9.197
(c) 10.197
(d) 11.197
Show Answer & Explanation

Answer: (c) 10.197

Explanation:
1. \( \ce{Na2CO3}=30/106=283 \) mmol; HCl converts 150 mmol to \( \ce{HCO3-} \), leaving 133 mmol \( \ce{CO3^2-} \).
2. \( pK_a=10.25 \); \( pH=10.25+\log\frac{133}{150}=10.20 \).

Teacher's Note:
1. The pair is carbonate and bicarbonate.

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Ionic Equilibrium Objective Questions & Solutions for JEE Chemistry

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