3. IONIC EQUILIBRIA Textbook Solution Class 12 Chemistry Maharashtra Board
1. Choose the most correct answer :
i. The pH of 10-8 M of HCl is
a. 8
b. 7
c. less than 7
d. greater than 7
Answer:-
i. The pH of 10-8 M of HCl is less than 7
ii. Which of the following solution will have pH value equal to 1.0 ?
a. 50 mL of 0.1M HCl + 50mL of 0.1M NaOH
b. 60 mL of 0.1M HCl + 40mL of 0.1M NaOH
c. 20 mL of 0.1M HCl + 80mL of 0.1M NaOH
d. 75 mL of 0.2M HCl + 25mLof 0.2M NaOH
Answer:-
d. 75 mL of 0.2M HCl + 25mLof 0.2M NaOH
iii. Which of the following is a buffer solution ?
a. CH3COONa + NaCl in water
b. CH3COOH + HCl in water
c. CH3COOH+CH3COONa in water
d. HCl + NH4Cl in water
Answer:-
c. CH3COOH+CH3COONa in water
iv. The solubility product of a sparingly soluble salt AX is 5.2×10-13. Its solubility in mol dm-3 is
a. 7.2 × 10-7
b. 1.35 × 10-4
c. 7.2 × 10-8
d. 13.5 × 10-8
The solubility product of a sparingly soluble salt \(AX\) is \(5.2 \times 10^{-13}\). Its solubility in mol dm\(^{-3}\) is \(\underline{7.2 \times 10^{-7}}\).
Explanation:
\(K_{sp} = 5.2 \times 10^{-13}\)
\[ \ce{AX <=> A^{-} + X^{+}} \]
\(K_{sp} = [A^{-}][X^{+}] = s \times s\)
\(5.2 \times 10^{-13} = s^{2}\)
\(\sqrt{10^{-13}} = \sqrt{10^{-12}} = 10^{-6}\)
\(0.52 \times 10^{-12} = s^{2}\)
\(\sqrt{0.52 \times 10^{-12}} = s\)
\(0.7211 \times 10^{-6} = s\)
\(s = 7.211 \times 10^{-7}\)
v. Blood in human body is highly buffered at pH of
a. 7.4
b. 7.0
c. 6.9
d. 8.1
Answer:-
Blood in human body is highly buffered at pH of
7.4
vi. The conjugate base of [Zn(H2O)4]2⊕ is
a. [Zn(H2O)4]2 NH3
b. [Zn(H2O)3]2
c. [Zn(H2O)3OH]⊕
d. [Zn(H2O)H]3⊕
The conjugate base of [Zn(H2O)4]2⊕ is [Zn(H2O)3OH]⊕
vii. For pH > 7 the hydronium ion concentration would be
a. 10-7M b. < 10-7M
c. > 10-7M d. ≥ 10-7M
Answer:- For pH > 7 the hydronium ion concentration would be < 10-7M
2. Answer the following in one sentence :
i. Why cations are Lewis acids ?.
Answer:- Cations are electron-deficient species and can accept an electron pair. Hence, cations are Lewis acids.
ii. Why is KCl solution neutral to litmus?.
Answer:-
KCl, being salt of a strong acid (HCl) and a strong base (KOH), does not undergo hydrolysis. Hence, the KCl solution is neutral to litmus.
iii. How are basic buffer solutions prepared?
Answer:-
1. Basic buffer solution is prepared by mixing aqueous solutions of a weak base like NH4OH and its salt of a strong acid like NH4Cl.
2. A weak base is selected according to the required pH or pOH of the solution and dissociation constant of the weak base.
iv. Dissociation constant of acetic acid is 1.8 × 10-5. Calculate percent dissociation of acetic acid in 0.01 M solution.
Given:
- Dissociation constant (\(K_a\)) = \(1.8 \times 10^{-5}\),
- Concentration (\(c\)) = 0.01 M
To find:
- Percent dissociation
Formulae:
- \(K_a = \alpha^2c\)
- Percent dissociation = \(\alpha \times 100\)
Calculation:
\(c = 0.01\) M = \(1 \times 10^{-2}\) M
Using formula (i),
Therefore, \(\alpha = \sqrt{\frac{K_a}{c}}\)
\(= \sqrt{\frac{1.8 \times 10^{-5}}{1 \times 10^{-2}}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{18 \times 10^{-4}}\)
\(= 4.242 \times 10^{-2}\)
Using formula (ii),
Percent dissociation \(= \alpha \times 100 = 4.242 \times 10^{-2} \times 100 = 4.242\%\)
Percent dissociation of 0.01 M acetic acid solution is 4.242%.
v. Write one property of a buffer solution.
Answer:-
i. by addition of small amount of either
strong acid or strong base,
ii. on dilution or
iii. when it is kept for long time.
vi. The pH of a solution is 6.06. Calculate its H⊕ ion concentration.
Given: pH of solution = 6.06
To find: \(H^+\) ion concentration
Formula: pH = \(-\log_{10}[H_3O^+]\)
Calculation: From the formula,
pH = \(-\log_{10}[H_3O^+]\)
Therefore, \(\log_{10}[H_3O^+] = -\text{pH}\)
= \(-6.06\)
= \(-6 - 0.06 + 1 - 1\)
= \((-6 - 1) + 1 - 0.06\)
= \(-7 + 0.94\)
= \(-6.06\)
Thus, \([H_3O^+] = \text{Antilog}_{10}[-6.06]\)
= \(8.710 \times 10^{-7}\) M
The \(H^+\) ion concentration of the solution is \(8.710 \times 10^{-7}\) M.
vii. Calculate the pH of 0.01 M sulphuric acid.
Given: Concentration of sulfuric acid = 0.01 M
To find: pH
Formula: pH = \(-\log_{10}[H_3O^+]\)
Calculation:
Sulfuric acid (\(H_2SO_4\)) is a strong acid. It dissociates almost completely in water as:
\[H_2SO_{4(aq)} + 2H_2O_{(l)} \rightarrow 2H_3O^+_{(aq)} + SO^{2-}_{4(aq)}\]
Hence, \([H_3O^+] = 2 \times c = 2 \times 0.01 \, \text{M} = 2 \times 10^{-2} \, \text{M}\)
From formula (i),
pH = \(-\log_{10}[H_3O^+] = -\log_{10}(2 \times 10^{-2}) = -\log_{10}2 - \log_{10}(10^{-2})\)
= \(-\log_{10}2 + 2 = 2 - 0.3010\)
pH = 1.699
The pH of 0.01 M sulfuric acid is 1.699.
viii. The dissociation of H2S is suppressed in the presence of HCl. Name the phenomenon
Answer:-
The phenomenon due to which dissociation of H2S is suppressed in the presence of HCl is known as the common ion effect
ix. Why is it necessary to add H2SO4 while preparing the solution of CuSO4?
i. The aqueous solution of CuSO4 is turbid due to the formation of sparingly soluble Cu(OH)2 by hydrolysis as shown below:
\[Cu^{2+}_{(aq)} + 4H_2O_{(l)} \rightleftharpoons Cu(OH)_{2(aq)} + 2H_3O^+_{(aq)}\]ii. If H2SO4 (i.e., H3O+ ions) are added, the hydrolytic equilibrium shifts to the left. Turbidity of Cu(OH)2 dissolves to give a clear solution. Hence, it is necessary to add H2SO4 while preparing the solution of CuSO4.
x. Classify the following buffers into different types :
a. CH3COOH + CH3COONa
b. NH4OH + NH4Cl
c. Sodium benzoate + benzoic acid
d. Cu(OH)2 + CuCl2
Answer:-
a. CH3COOH + CH3COONa
Acidic Buffer
b. NH4OH + NH4Cl
Basic Buffer
c. Sodium benzoate + benzoic acid
Acidic Buffer
d. Cu(OH)2 + CuCl2
Basic Buffer
3. Answer the following in brief :
i. What are acids and bases according to Arrhenius theory ?
According to Arrhenius theory, acids and bases are defined as follows:
i. Acid: An acid is a substance that contains hydrogen and gives H+ ions in an aqueous solution.
e.g.
\[\ce{HCl_{(aq)} ->[water]H^+_{ (aq)} + Cl^-_{ (aq)}}\];
\[\ce{CH3COOH_{(aq)}⇌[water] H^+_{ (aq)} + CH3COO^-_{ (aq)}}\]
ii. Base: A base is a substance that contains the OH group and produces hydroxide ions (`"OH"^-` ions) in aqueous solution.
e.g.
\[\ce{NaOH_{(aq)}->[water]Na^+_{ (aq)} + OH^-_{ (aq)}}\];
\[\ce{NH4OH_{(aq)} ⇌[water] NH^+_{ 4(aq)} + OH^-_{ (aq)}}\]
ii. What is meant by conjugate acidbase pair?
Answer:-
Conjugate acid-base pair : A pair of an acid and a base differing by a proton is called a conjugate acid-base pair.
iii. Label the conjugate acid-base pair in the following reactions
a. HCl + H2O H3O⊕ + Cl
b. CO3 2 + H2O OH + HCO3
\[\ce{HCl + H2O ⇌ H3O+ + Cl-}\]
\[\ce{CO^2-_3 + H2O ⇌ OH- + HCO^-_3}\]
Answer:-
\[\ce{\underset{Acid1}{HCl} + \underset{Base2}{H2O} ⇌ \underset{Acid2}{H3O+} + \underset{Base1}{Cl-}}\]
\[\ce{\underset{Base1}{CO^2-_3} + \underset{Acid2}{H2O} ⇌ \underset{Base2}{OH-} + \underset{Acid1}{HCO^-_3}}\]
iv. Write a reaction in which water acts as a base.
Answer:-
\[\ce{\underset{(Base)}{H2O}_{(l)} + HCl_{(aq)} ⇌ H3O^+_{ (aq)} + Cl^-_{ (aq)}}\]
v. Ammonia serves as a Lewis base whereas AlCl3 is Lewis acid. Explain.
Answer:-
Since ammonia molecule, NH3 has a lone pair of electrons to donate it acts as a Lewis base. AlCl3 is a molecule with incomplete octet hence it is electron deficient and acts as a Lewis acid.
vi. Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid
Given: Percent dissociation = 5%, Concentration (c) = 1 decimolar
To find: Dissociation constant of acid (\(K_a\))
Formulae:
i. Percent dissociation = \(\alpha \times 100\)
ii. \(K_a = \alpha^2c\)
Calculation:
Using formula (i),
\(\alpha = \frac{{\text{Percent dissociation}}}{{100}} = \frac{5}{100} = 0.05\)
c = 1 decimolar = 0.1 M
Using formula (ii),
\(K_a = (0.05)^2 \times 0.1\)
= \(2.5 \times 10^{-4}\)
The dissociation constant of the acid is \(2.5 \times 10^{-4}\).
vii. Derive the relation pH + pOH = 14.
Relationship between pH and pOH:
The ionic product of water is given as:
\[K_w = [H_3O^+][OH^-]\]Now, \(K_w = 1 \times 10^{-14}\) at 298 K
Thus, \([H_3O^+][OH^-] = 1.0 \times 10^{-14}\)
Taking the logarithm of both sides, we write
\[ \log_{10}[H_3O^+] + \log_{10}[OH^-] = -14 \] \[ -\log_{10}[H_3O^+] + (-\log_{10}[OH^-]) = 14 \]Now, pH = -\log_{10}[H_3O^+] and pOH = -\log_{10}[OH^-]
Therefore, \(pH + pOH = 14\)
viii. Aqueous solution of sodium carbonate is alkaline whereas aqueous solution of ammonium chloride is acidic. Explain.
i. Sodium carbonate (\(Na_2CO_3\)) is a salt of weak acid \(H_2CO_3\) and strong base \(NaOH\). When dissolved in water, it dissociates completely.
\[\ce{Na2CO3_{(aq)} -> 2Na^+_{(aq)} + CO3^{2-}_{(aq)}}\]ii. The \(Na^+\) ions of salt have no tendency to react with \(OH^-\) ions of water since the possible product of the reaction is \(NaOH\), a strong electrolyte.
iii. On the other hand, the reaction of \(CO_3^{2-}\) ions of salt with the \(H_3O^+\) ions from water produces unionized \(H_2CO_3\).
\[\ce{CO3^{2-}_{(aq)} + 2H2O_{(l)} <-> H2CO3_{(aq)} + 2OH^-_{(aq)}}\]Thus, the hydrolytic equilibrium for \(Na_2CO_3\) is,
\[\ce{Na2CO3_{(aq)} + H2O_{(l)} <-> H2CO3_{(aq)} + 2Na^+_{(aq)} + 2OH^-_{(aq)}}\]iv. As a result of excess \(OH^-\) ions produced, the resulting solution of \(Na_2CO_3\) is alkaline.
v. Similarly, ammonium chloride (\(NH_4Cl\)) is the salt of strong acid \(HCl\) and weak base \(NH_4OH\). When \(NH_4Cl\) is dissolved in water, it dissociates completely as,
\[\ce{NH4Cl_{(aq)} -> NH4^+_{(aq)} + Cl^-_{(aq)}}\]vi. \(Cl^-\) ions of salt have no tendency to react with water because the possible product \(HCl\) is a strong electrolyte.
vii. The reaction of \(NH_4^+\) ions with \(OH^-\) ions forms unionized \(NH_4OH\). The hydrolytic equilibrium for \(NH_4Cl\) is then written as,
\[\ce{NH4^+_{(aq)} + 2H2O_{(l)} <-> NH4OH_{(aq)} + H3O+_{(aq)}}\]viii. Due to the presence of an excess of \(H_3O^+\) ions, the resulting solution of \(NH_4Cl\) is acidic.