Salt Hydrolysis Questions

The neutralisation reaction between an acid and a base produces salt. In general, they ionise in water, producing cations and anions. The cations and anions formed during salt ionisation either exist in aqueous solutions as hydrated ions or interact with water to regenerate the acids and bases. The interaction of salt anions or cations with water is referred to as salt hydrolysis.

Definition: It is defined as the reaction of a salt with water to produce acid and base.

Salt Hydrolysis Chemistry Questions with Solutions

Q-1: Which of the following statements about the degree of hydrolysis for strong base and weak acid is incorrect?

  1. It decreases with dilution
  2. It is dependent of dilution
  3. It increases with increase in temperature
  4. It increases with decrease in Ka of the bases

Answer: a) It decreases with dilution

Explanation: We know that for strong base and weak acid, degree of hydrolysis is given by

\(\begin{array}{l}h=\sqrt{\frac{K_{W}}{c\times K_{a}}}\end{array} \)

As a result of dilution, concentration decreases. The relationship clearly shows that the degree of hydrolysis is inversely proportional to concentration. As a result, as concentration decreases, the degree of hydrolysis increases. This means it gets stronger with dilution.

Q-2: Which of the salts undergoes cationic hydrolysis?

  1. NaH2PO4
  2. CH3COONa
  3. GaCl3
  4. Li2CO3

Answer: C) GaCl3

Explanation: Cationic hydrolysis occurs in salts of strong acids and weak bases. Because GaCl3 is a salt of both a weak base (Ga(OH)3) and a strong acid (HCl), it will undergo cationic hydrolysis. Rest salts are salts of a strong base and a weak acid, so they will not undergo this type of hydrolysis.

Q-3: In which case both pH and hydrolysis constant are independent of the concentration of ions involved in hydrolysis?

  1. WA+SB
  2. SA+WB
  3. WA+WB
  4. SA+SB

Answer: c) WA+WB

Explanation: For a salt of weak acid(WA) and weak base(WB), the expression for pH is

pH = ½(pKW+pKa-pKb)

And for hydrolysis constant, is

Kh = KW/(Ka×Kb)

We can clearly see that neither the hydrolysis constant nor the pH expressions contain concentration terms. This means that it is unaffected by the concentration of an ion involved in hydrolysis.

Q-4: The degree of hydrolysis of a salt of a weak acid and a weak base in 0.1M solution is found to be 30%. If the molarity of the solution is 0.2 M, the salt’s percent hydrolysis should be

  1. 90%
  2. 30%
  3. 65%
  4. 10%

Answer: b) 30%

Explanation: The degree of hydrolysis of a weak acid and weak salt is independent of the concentration of the solution. Hence, when the molarity of the solution is 0.2 M, the percentage degree of hydrolysis will remain the same,that is, 30%.

Q-5: At 25 degrees Celsius, the sodium salt of a weak monobasic organic acid is hydrolyzed to a 3 percent extent in its 0.1M solution. Given that the ionic product of water at this temperature is 10-14, what is the acid’s dissociation constant?

Answer: Given % hydrolysis,h = 3% = 0.03

For a sodium salt of weak acid, formula for acid’s dissociation constant(Ka) is

Ka = KW/Kh

Where KW is the ionic product of water and Kh is the hydrolysis constant.

Kh= h2c = 0.032 ×0.1 = 9×10-5

Substitute Kw = 10-14 and Kh=9×10-5.in the formula for Ka

Ka = KW/Kh

Ka = 10-14 / 9×10-5

= 10-10( approximately)

Q-6: Calculate the % hydrolysis in a 10-2 M solution of NaCN (Ka = 6.2×10-10).

Answer: NaCN is a salt of weak acid(HCN) and strong base(NaOH). The degree of hydrolysis(h) for such as salt can be calculated by using the following formulae:

\(\begin{array}{l}h= \sqrt{\frac{K_{W}}{K_{ a}\times C}}\end{array} \)
\(\begin{array}{l}h= \sqrt{\frac{10^{-14}}{6.2\times 10^{-10}\times 10^{-2}}} = 0.0401\end{array} \)

Thus, % hydrolysis = %h = 0.0401×100 = 4.0%

Q-7: NH4Cl has a hydrolysis constant (Kh) of 5.6×10-10.Calculate the pH at equilibrium for 0.1 M NH4Cl solution.

Answer: NH4Cl is a salt of weak base (NH4OH) and strong acid (HCl). The hydrolysis constant for such a salt has the following formulae:

\(\begin{array}{l}K_{h}=\frac{K_{w}}{K_{b}}\end{array} \)

Substitute Kw = 10-14 and Kh= 5.6×10-10 in the above formula and calculate for Kb.

\(\begin{array}{l}5.6\times 10^{-10}=\frac{10^{-14}}{K_{b}}\end{array} \)

On solving, Kb= 1.78 × 10-5

pKb= -log(Kb)= 4.74

The pH for this salt can be calculated using the below equation:

pH = ½[pKw-pKb-log c] =½ [14-4.74-log(0.1)] = 5.13

Q-8: Salt hydrolysis is a chemical reaction that occurs when salt reacts with _______

  1. Acid
  2. Base
  3. Water
  4. Salt

Answer: c) Water

Q-9: AlBr3 is formed by the reaction of a strong acid and a weak base. What is the nature of the solution,when this salt dissociates?

  1. Basic
  2. Neutral
  3. Basic and acidic
  4. Acidic

Answer: d) Acidic

Q-10: Give the hydrolysis reaction for magnesium nitride.

Answer: The hydrolysis reaction for magnesium nitride(Mg3N2) is shown below:

\(\begin{array}{l}Mg_{3}N_{2}+ 6H_{2}O\to 3Mg(OH)_{2}+2NH_{3}\end{array} \)

Q-11: Which nitrogen oxides are formed on the hydrolysis of N2O3?

  1. N2O + NO
  2. NO2 + N2O
  3. N2O + NO2
  4. None of the above

Answer: A) N2O + NO

Q-12: The compound with highest pH among the following is

  1. CH3COONH4
  2. Na2CO3
  3. NH4Cl
  4. NaNO3

Answer: b) Na2CO3

Explanation: The basic salt has the highest pH. Na2CO3 is the salt of NaOH (a strong base) and H2CO3 (weak acid). Overall, the salt will be basic, resulting in the highest pH.

Q-13: Kb for NH4OH is 1.8 ×10-5 and Ka for CH3COOH is 1.8×10-5. Ammonium acetate’s pH will be

  1. 7.08
  2. 6.75
  3. 7.0
  4. Between 6 and 7

Answer: c) 7.0

Explanation: Ammonium acetate is a salt of weak acid and weak base. For a salt of weak acid(WA) and weak base(WB), pH can be calculated using the below formula:

pH = ½(pKW+pKa-pKb)

We can see Kb and Ka values are the same, it means their pKa and pKb values are also the same.

pH= ½(pKW) = 14/2 = 7

Q-14: The pH of a 0.02 M solution of pyridinium hydrochloride is 3.44. Calculate the pyridine’s ionisation constant.

Answer: Pyridinium hydrochloride is a weak base and strong acid(HCl) salt. We need to calculate ionisation constant(Kb).

The pH formula for this salt is shown below:

pH = ½[pKw-pKb-log c]

3.44= ½[14-pKb-log(0.02)]

pKb.= 8.82

pKb = -logKb

Kb = 1.51 × 10-9

Q-15: The hydrolysis of weak acid and strong base is called

  1. Cationic hydrolysis
  2. Anionic hydrolysis
  3. Non ionic hydrolysis
  4. Amphoteric hydrolysis

Answer: b) Anionic hydrolysis

Practise Questions on Salt Hydrolysis

Q-1: HX is a weak acid( Ka= 10-5).When it reacts with caustic soda, it forms the salt NaX(0.1 M). The degree of hydrolysis of NaX is

  1. 0.1%
  2. 0.000001%
  3. 0.01%
  4. 0.5%

Q-2: Determine whether the following salts’ aqueous solutions are acidic, basic, or neutral.

  1. NH4F
  2. Na2HPO4
  3. NH4Cl

Q-3: A salt hydrolysis reaction occurs when

  1. Salts dissociate in any liquid solvent to produce proton and hydroxide ions.
  2. Salt dissociates in water to produce acidic and basic solutions.
  3. The salt releases anions and cations, resulting in acidic solutions.
  4. The salt releases anions and cations, resulting in basic solutions.

Q-4: Which of the following salt’s degree of hydrolysis is independent of solution concentration?

  1. NH4CN
  2. CH3COONa
  3. (NH4)2SO4
  4. None of these

Q-5: The pH of 0.4M aqueous NaCN solution will be (Given pKb of CN– = 4.70)

  1. 2.548
  2. 3.513
  3. 10.213
  4. 11.450

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