JEE Chemistry Ionic Equilibrium MCQs Set 19

Chemistry Objective Questions and Answers: Ionic Equilibrium

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Multiple Choice Questions

Question 1: The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt BA will be
(a) 8.58
(b) 4.79
(c) 7.01
(d) 9.22
Show Answer & Explanation

Answer: (c) 7.01

Explanation:
1. For the salt of a weak acid and a weak base, \( pH = 7 + \frac{1}{2}(pK_a - pK_b) \).
2. pH = 7 + 0.5(4.80 - 4.78) = 7.01.

Teacher's Note:
1. The pH is near 7 when the acid and base are about equally weak.
2. The result is independent of concentration.

Question 2: The pH of a K2S solution is
(a) 7
(b) less than 7
(c) more than 7
(d) 0
Show Answer & Explanation

Answer: (c) more than 7

Explanation:
1. K2S is the salt of a strong base (KOH) and a weak acid (H2S).
2. S2- hydrolyses as S2- + H2O gives HS- + OH-, producing a basic solution with pH above 7.

Teacher's Note:
1. Anions of weak acids hydrolyse to give OH-.
2. K+ does not hydrolyse.

Question 3: If 40 ml of 0.2 M KOH is added to 160 ml of 0.1 M HCOOH (Ka = \( 2 \times 10^{-4} \)), the pOH of the resulting solution is
(a) 3.4
(b) 3.7
(c) 7
(d) 10.3
Show Answer & Explanation

Answer: (d) 10.3

Explanation:
1. KOH = 8 mmol and HCOOH = 16 mmol, so 8 mmol HCOOH remains and 8 mmol HCOO- forms.
2. This is a buffer with equal acid and salt, so pH = pKa = 3.7.
3. pOH = 14 - 3.7 = 10.3.

Teacher's Note:
1. Half neutralisation of a weak acid gives pH = pKa.
2. Convert to pOH at the end.

Question 4: 1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is
(a) not a buffer solution and with pH less than 7
(b) not a buffer solution and with pH more than 7
(c) a buffer solution with pH less than 7
(d) a buffer solution with pH more than 7
Show Answer & Explanation

Answer: (a) not a buffer solution and with pH less than 7

Explanation:
1. A buffer needs a weak acid and its salt (or a weak base and its salt).
2. HCl is a strong acid and NaCl is its neutral salt, so there is no buffer action.
3. The solution has [H+] = 1 M, so the pH is less than 7.

Teacher's Note:
1. Strong acid with its salt is not a buffer.
2. Buffers need a weak conjugate pair.

Question 5: The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionised is
(a) 4.5
(b) 2.5
(c) 9.5
(d) 7.0
Show Answer & Explanation

Answer: (c) 9.5

Explanation:
1. If 50% is ionised then [HA] = [A-].
2. By Henderson's equation pH = pKa = 4.5.
3. pOH = 14 - 4.5 = 9.5.

Teacher's Note:
1. At half ionisation pH = pKa.
2. Remember to convert to pOH.

Question 6: To 50 ml of 0.05 M formic acid, what volume of 0.10 M sodium formate must be added to get a buffer solution of pH = 4.0? (pKa of the acid is 3.7, log 2 = 0.3)
(a) 40 ml
(b) 4 ml
(c) 50 ml
(d) 100 ml
Show Answer & Explanation

Answer: (c) 50 ml

Explanation:
1. pH = pKa + log (salt / acid): 4.0 = 3.7 + log(salt/acid), so salt/acid = 2.
2. Acid = 50 x 0.05 = 2.5 mmol, so salt = 5 mmol.
3. Volume of 0.10 M sodium formate = 5/0.10 = 50 ml.

Teacher's Note:
1. Use Henderson's equation with moles.
2. Both acid and salt are in the same final volume.

Question 7: Which can act as a buffer?
(a) NH4OH + NaOH
(b) HCOOH + HCl
(c) 40 ml of 0.1 M NaCN + 20 ml of 0.1 M HCl
(d) all of them
Show Answer & Explanation

Answer: (c) 40 ml of 0.1 M NaCN + 20 ml of 0.1 M HCl

Explanation:
1. In (a) and (b) a strong base or strong acid is mixed with a weak one, with no conjugate pair formed.
2. In (c) 4 mmol CN- reacts with 2 mmol HCl to give 2 mmol HCN, leaving 2 mmol CN-. This is a weak acid with its conjugate base, so it is a buffer.

Teacher's Note:
1. A buffer is a weak acid with its salt.
2. A partly neutralised weak base or acid forms a buffer.

Question 8: If equal volumes of 0.05 M ammonium hydroxide solution and 0.001 M ammonium chloride solution are mixed, what will be the OH- ion concentration of the solution? Kb(NH4OH) = \( 1.8 \times 10^{-5} \)
(a) \( 3.0 \times 10^{-3} \)
(b) \( 4.6 \times 10^{-4} \)
(c) \( 9.0 \times 10^{-3} \)
(d) \( 9.0 \times 10^{-4} \)
Show Answer & Explanation

Answer: (b) \( 4.6 \times 10^{-4} \)

Explanation:
1. After mixing, [NH4OH] = 0.025 M and [NH4+] = 0.0005 M.
2. Let [OH-] = x. Then x(0.0005 + x)/(0.025 - x) = \( 1.8 \times 10^{-5} \).
3. The approximation of using the salt concentration only is not valid here, so solve \( x^{2} \) + 0.0005x - \( 4.5 \times 10^{-7} \) = 0, giving x = \( 4.6 \times 10^{-4} \) M.

Teacher's Note:
1. The usual buffer approximation needs the salt concentration to be much larger than x.
2. Here the quadratic must be solved.

Question 9: Calculate the pH of a buffer made by mixing 600 cc of 0.6 M NH3 and 400 cc of 0.5 M NH4Cl. Kb for NH3 = \( 1.8 \times 10^{-5} \) (log 1.8 = 0.26)
(a) 11.3
(b) 9.0
(c) 9.52
(d) 5
Show Answer & Explanation

Answer: (c) 9.52

Explanation:
1. NH3 = 360 mmol and NH4Cl = 200 mmol.
2. pKb = 5 - 0.26 = 4.74.
3. pOH = 4.74 + log(200/360) = 4.74 - 0.26 = 4.48, so pH = 14 - 4.48 = 9.52.

Teacher's Note:
1. For a basic buffer pOH = pKb + log (salt / base).
2. Use moles, the volume cancels out.

Question 10: On addition of NaOH to a CH3COOH solution, 60% of the acid is neutralised. If pKa of CH3COOH is 4.7 then 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) remains unchanged
Show Answer & Explanation

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

Explanation:
1. [salt]/[acid] = 60/40 = 1.5.
2. pH = 4.7 + log 1.5 = 4.7 + 0.18 = 4.88, which lies between 4.7 and 5.0.

Teacher's Note:
1. More than half neutralised means pH above pKa.
2. Use the ratio of salt to remaining acid.

Question 11: What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole of propanoic acid (Ka = \( 3 \times 10^{-5} \) at 25 C) to obtain a buffer solution of pH 4.7?
(a) \( 4.52 \times 10^{-2} \) mol
(b) \( 3.52 \times 10^{-2} \) mol
(c) \( 2.52 \times 10^{-2} \) mol
(d) \( 3 \times 10^{-2} \) mol
Show Answer & Explanation

Answer: (d) \( 3 \times 10^{-2} \) mol

Explanation:
1. [H+] = \( 10^{-4.7} \) = \( 2 \times 10^{-5} \) M.
2. From Ka = [H+][salt]/[acid]: salt = Ka x acid / [H+] = \( 3 \times 10^{-5} \) x 0.02 / (\( 2 \times 10^{-5} \)) = 0.03 mol.

Teacher's Note:
1. Rearrange the Ka expression to find the salt.
2. \( 10^{-4.7} \) is about \( 2 \times 10^{-5} \).

Question 12: In a buffer solution the ratio of concentration of NH4Cl and NH4OH is 1 : 1. When it changes to 2 : 1, what will be the change in pH of the buffer?
(a) increases
(b) decreases
(c) no effect
(d) none of these
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. pOH = pKb + log([NH4Cl]/[NH4OH]).
2. When the ratio goes from 1 to 2, pOH increases by log 2.
3. So [OH-] falls and the pH decreases.

Teacher's Note:
1. More salt in a basic buffer lowers the pH.
2. The pH changes by only 0.3 units.

Question 13: The buffer solution plays an important role in
(a) increasing the pH value
(b) decreasing the pH value
(c) keeping the pH constant
(d) making the solution neutral
Show Answer & Explanation

Answer: (c) keeping the pH constant

Explanation:
1. A buffer resists changes in pH when small amounts of acid or base are added.
2. So it keeps the pH nearly constant.

Teacher's Note:
1. Buffers are used in blood, in analysis and in industry.
2. They do not make a solution neutral.

Question 14: Half of the formic acid solution is neutralised on addition of a KOH solution to it. If Ka (HCOOH) = \( 2 \times 10^{-4} \), then 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. Half neutralisation gives [HCOOH] = [HCOO-], so pH = pKa.
2. pKa = -log(\( 2 \times 10^{-4} \)) = 4 - 0.3010 = 3.6990.

Teacher's Note:
1. At the half equivalence point pH = pKa.
2. Use the given logarithm.

Question 15: When 0.02 moles of NaOH are added to a litre of buffer solution, its pH changes from 5.75 to 5.80. What is its buffer capacity?
(a) 0.4
(b) 0.05
(c) -0.05
(d) 2.5
Show Answer & Explanation

Answer: (a) 0.4

Explanation:
1. Buffer capacity is the moles of base added per litre divided by the change in pH.
2. Capacity = 0.02 / (5.80 - 5.75) = 0.02/0.05 = 0.4.

Teacher's Note:
1. A larger value means the buffer resists change better.
2. Use the change in pH as a positive number.

Question 16: Calculate the pH when 100 ml of 0.2 M NaOH is reacted with 100 ml of 0.2 M CH3COOH (Ka = \( 10^{-5} \)).
(a) 9
(b) 7
(c) 5
(d) 2
Show Answer & Explanation

Answer: (a) 9

Explanation:
1. Equal moles react completely to give 20 mmol of CH3COONa in 200 ml, which is 0.1 M.
2. For the salt of a weak acid pH = 7 + (pKa + log C)/2 = 7 + (5 - 1)/2 = 9.

Teacher's Note:
1. At the equivalence point of a weak acid with a strong base the solution is basic.
2. Use the salt hydrolysis formula.

Question 17: At 90 C pure water has [H+] = \( 10^{-6} \) M. If 100 ml of 0.2 M HNO3 is added to 20 ml of 1 M NaOH at 90 C, then the pH of the resulting solution will be
(a) 5
(b) 6
(c) 7
(d) none of these
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. HNO3 = 20 mmol and NaOH = 20 mmol, so they neutralise exactly.
2. The resulting solution is neutral at 90 C, where [H+] = \( 10^{-6} \) M.
3. So pH = 6.

Teacher's Note:
1. Neutral pH depends on temperature.
2. At 90 C Kw = \( 10^{-12} \).

Question 18: When 20 ml of M/20 NaOH is added to 10 ml of M/10 HCl, the resulting solution will
(a) turn blue litmus red
(b) turn phenolphthalein solution pink
(c) turn methyl orange red
(d) have no effect on either red or blue litmus
Show Answer & Explanation

Answer: (d) have no effect on either red or blue litmus

Explanation:
1. NaOH = 20 x 1/20 = 1 mmol and HCl = 10 x 1/10 = 1 mmol.
2. They neutralise exactly, giving a neutral solution of NaCl.
3. So there is no effect on litmus.

Teacher's Note:
1. Calculate moles before judging acidic or basic.
2. Equal moles of strong acid and base give a neutral salt.

Question 19: The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. The pH of the solution is related to the ratio of the concentrations of the conjugate acid (HIn) and base (In-) forms of the indicator by the expression
(a) log [HIn]/[In-] = pKIn - pH
(b) log [HIn]/[In-] = pH - pKIn
(c) log [In-]/[HIn] = pH + pKIn
(d) log [In-]/[HIn] = pKIn - pH
Show Answer & Explanation

Answer: (a) log [HIn]/[In-] = pKIn - pH

Explanation:
1. For an indicator HIn going to H+ + In-, pH = pKIn + log([In-]/[HIn]).
2. So log([HIn]/[In-]) = pKIn - pH.

Teacher's Note:
1. This is Henderson's equation for the indicator.
2. Flip the ratio and the sign changes.

Question 20: Calculate the pH range in which an acid indicator with Ka (indicator) = \( 1.0 \times 10^{-5} \) changes colour when the concentration of the indicator is \( 1 \times 10^{-3} \) M.
(a) 5 +/- 1
(b) 11 +/- 1
(c) 3 +/- 1
(d) 8 +/- 1
Show Answer & Explanation

Answer: (a) 5 +/- 1

Explanation:
1. An indicator changes colour in the range pKa +/- 1.
2. pKa = 5, so the range is 5 +/- 1.

Teacher's Note:
1. The range is where the ratio of the two forms goes from 1:10 to 10:1.
2. It does not depend on the indicator concentration.

Question 21: Phenolphthalein is a
(a) strong acid
(b) strong base
(c) weak base
(d) weak acid
Show Answer & Explanation

Answer: (d) weak acid

Explanation:
1. Phenolphthalein is a weak organic acid.
2. It is colourless in its unionised form and pink in the ionised form, which forms in basic medium.

Teacher's Note:
1. Most acid-base indicators are weak acids or weak bases.
2. The ionised form of phenolphthalein is pink.

Question 22: The pH range of the methyl red indicator is
(a) 4.2 - 6.2
(b) 6.8 - 10.8
(c) 8 - 9.6
(d) 6.8 - 8.2
Show Answer & Explanation

Answer: (a) 4.2 - 6.2

Explanation:
1. Methyl red changes colour from red to yellow between pH 4.2 and 6.2.
2. The other ranges belong to other indicators such as phenolphthalein (8.3 to 10) and phenol red.

Teacher's Note:
1. Learn the range of common indicators.
2. Methyl orange is 3.1 to 4.4.

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Multiple Choice Questions (MCQs) for JEE Chemistry Ionic Equilibrium

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