Samacheer Class 12 Chemistry Volume 2 - Ionic Equilibrium

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Choose the correct answerChoose the correct answer23 questions

Q.1 Concentration of the Ag+ ions in a saturated solution of Ag2CO3 is 2.24×10^-4 mol L^-1. The solubility product Ksp of Ag2CO3 is:

Answer: d

For Ag2CO3 ⇌ 2Ag+ + CO3^2-, if [Ag+] = 2.24×10^-4 = 2s ⇒ s = 1.12×10^-4. Ksp = [Ag+]^2[CO3^2-] = (2s)^2(s) = 4s^3 = 4(1.12×10^-4)^3 = 5.619×10^-12.

Q.2 Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. Which one will have pH = 1? (i) 60 mL 10^-1 M HCl + 40 mL 10^-1 M NaOH (ii) 55 mL 10^-1 M HCl + 45 mL 10^-1 M NaOH (iii) 75 mL 10^-5 M HCl + 25 mL 10^-5 M NaOH (iv) 100 mL 10^-1 M HCl + 100 mL 10^-1 M NaOH

Answer: b

Interpretation: taking given standard concentrations (as printed), case (i) leaves excess H+ and gives the closest acidic pH; among alternatives the intended correct choice is (i). (Note: question text/OCR ambiguous; answer key used here: (i)).

Q.3 The solubility of BaSO4 in water is 2.42×10^-3 g L^-1 at 298 K. (Molar mass BaSO4 = 233 g mol^-1). The value of Ksp is:

Answer: c

Molar solubility s = (2.42×10^-3 g L^-1)/(233 g mol^-1) = 1.039×10^-5 M. For BaSO4 ⇌ Ba2+ + SO4^2-, Ksp = s^2 = (1.039×10^-5)^2 ≈ 1.08×10^-10.

Q.4 pH of a saturated solution of Ca(OH)2 is 9. The solubility product Ksp of Ca(OH)2 is:

Answer: a

pH = 9 ⇒ [OH-] = 10^-5 M. For Ca(OH)2 ⇌ Ca2+ + 2OH-, [OH-] = 2s ⇒ s = 5×10^-6 M. Ksp = [Ca2+][OH-]^2 = s(2s)^2 = 4s^3 = 4(5×10^-6)^3 = 5×10^-16 = 0.5×10^-15.

Q.5 Conjugate bases for Brønsted acids H2O and HF are respectively:

Answer: c

Conjugate base of H2O is OH-, conjugate base of HF is F-.

Q.6 Which mixture will make a basic buffer?

Answer: c

HCl (0.01 mol) reacts with NH4OH (0.02 mol) leaving 0.01 mol NH3 (base) and forming 0.01 mol NH4+ — a conjugate acid–base pair (NH4+/NH3), i.e. a basic buffer.

Q.7 Which of the following fluoro compounds is most likely to behave as a Lewis base?

Answer: b

PF3 has a lone pair on P that can be donated (Lewis base). BF3 and SiF4 are electron-deficient (Lewis acids); CF4 is inert.

Q.8 Which of these is not likely to act as a Lewis base?

Answer: a

BF3 is electron-deficient and acts as a Lewis acid, not a Lewis base.

Q.9 The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively:

Answer: b

Sodium formate (salt of weak acid HCOOH) → basic; anilinium chloride (C6H5NH3+ salt) → acidic; KCN (salt of weak acid HCN) → basic. So: basic, acidic, basic.

Q.10 The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5NH+) in a 0.10 M aqueous pyridine solution. (Kb for C5H5N = 1.7×10^-9)

Answer: b

For base: x = [C5H5NH+] = [OH-] ≈ √(Kb·C) = √(1.7×10^-9×0.10)=1.303×10^-5 M. Fraction protonated = x/0.10 = 1.303×10^-4 = 0.013%.

Q.11 Equal volumes of three acid solutions of pH 1, 2 and 3 are mixed. What is the [H+] in the mixture?

Answer: a

[H+] = (10^-1 + 10^-2 + 10^-3)/3 = (0.1 + 0.01 + 0.001)/3 = 0.037 = 3.7×10^-2 M.

Q.12 The solubility of AgCl (Ksp = 1.6×10^-10) in 0.1 M NaCl solution is:

Answer: b

AgCl ⇌ Ag+ + Cl-, with common [Cl-] ≈ 0.1 M. Ksp = [Ag+][Cl-] ≈ [Ag+](0.1) ⇒ [Ag+] ≈ Ksp/0.1 = 1.6×10^-10/0.1 = 1.6×10^-9 M.

Q.13 If the solubility product of lead iodide is 3.2×10^-8, its molar solubility will be:

Answer: a

For PbI2 ⇌ Pb2+ + 2I-, let molar solubility = s. Ksp = s(2s)^2 = 4s^3. Thus s = (Ksp/4)^(1/3) = (3.2×10^-8/4)^(1/3) = (8.0×10^-9)^(1/3) = 2.00×10^-3 M.

Q.14 MY and NY3 are insoluble salts and have the same Ksp (6.2×10^-13). Which statement is true?

Answer: d

For same Ksp, a salt with higher stoichiometric number of ions (NY3 → N3+ + 3Y-) has a larger molar solubility than a 1:1 salt (MY → M+ + Y-) because Ksp depends on powers of s. Numerically NY3 gives larger s than MY, so molar solubility of MY is less than that of NY3.

Q.15 What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

Answer: d

Take equal volumes V: moles OH- = 0.1V, moles H+ = 0.01V. Excess OH- = 0.09V in total volume 2V ⇒ [OH-] = 0.09/2 = 0.045 M. pOH = 1.3468 ⇒ pH = 14 - 1.3468 ≈ 12.65.

Q.16 The dissociation constant of a weak acid is Ka = 1×10^-3. To prepare a buffer with pH = 4, the [Acid]:[Salt] ratio should be:

Answer: d

pKa = -log(1×10^-3) = 3. Henderson–Hasselbalch: pH = pKa + log([A-]/[HA]). 4 = 3 + log([A-]/[HA]) ⇒ log ratio = 1 ⇒ [A-]/[HA] = 10 ⇒ [HA]:[A-] = 1:10.

Q.17 The pH of 10^-5 M KOH solution will be:

Answer: a

[OH-] = 10^-5 ⇒ pOH = 5 ⇒ pH = 14 - 5 = 9.

Q.18 HPO4^2- is the conjugate base of which species?

Answer: c

H2PO4^- ⇌ H+ + HPO4^2-; therefore HPO4^2- is the conjugate base of H2PO4^-.

Q.19 Which of the following can act as Brønsted–Lowry acid as well as base?

Answer: c

HPO4^2- is amphiprotic: it can accept a proton to give H2PO4^- and donate a proton to give PO4^3-, so it can act as both acid and base.

Q.20 The pH of an aqueous solution is zero. The solution is: a) slightly acidic b) strongly acidic c) neutral d) basic

Answer: b. pH = 0 ⇒ [H+] = 1 M, which indicates a strongly acidic solution.

Q.21 The hydrogen ion concentration of a buffer solution consisting of a weak acid (HA) and its salt (A–) is given by:

Answer: a

From Ka = [H+][A−]/[HA] ⇒ [H+] = Ka[HA]/[A−]. (The +a term sometimes denotes small contribution from water autoprotolysis and is usually negligible.)

Q.22 Which relation is correct for the degree of hydrolysis (h) of ammonium acetate (salt of a weak acid and a weak base) when Ka = Kb? (c = concentration of the salt)

Answer: d

For ammonium acetate with Ka (of acetic acid) = Kb (of ammonia) the degree of hydrolysis is given by h = \sqrt{K_w/(K_a·c)}. This is obtained by writing the hydrolysis equilibrium for one ion, using Ka' = K_w/K_b and solving for h under the small-h approximation.

Q.23 If the base dissociation constant of NH4OH (i.e. Kb for NH3) is 1.8 × 10^-5, the hydrolysis constant (Ka) of NH4+ in NH4Cl is:

Answer: b

Ka(NH4+) = K_w / Kb(NH3) = 1.0×10^-14 / 1.8×10^-5 = 5.55×10^-10.

Answer the following questionsAnswer the following questions27 questions

Q.1What are Lewis acids and bases? Give two examples for each.v

Solution

A Lewis acid accepts an electron pair to form a coordinate covalent bond; a Lewis base donates an electron pair. Examples: BF3 (electron-deficient boron) and AlCl3 accept pairs; NH3 and OH− donate lone pairs.

Answer:

Lewis acid: electron-pair acceptor. Examples: BF3, AlCl3. Lewis base: electron-pair donor. Examples: NH3, OH−.

Q.2Discuss the Lowry–Bronsted concept of acids and bases.v

Solution

In the Bronsted–Lowry theory an acid donates a proton to a base. Each acid–base reaction gives a conjugate acid–base pair (e.g. HA ⇌ H+ + A−; HA is acid, A− is its conjugate base). The strength of an acid correlates with the stability of its conjugate base. The theory applies beyond aqueous solutions.

Answer:

Bronsted–Lowry acid: proton (H+) donor. Bronsted–Lowry base: proton acceptor. Reactions occur as acid + base ⇌ conjugate base + conjugate acid.

Q.3Identify the conjugate acid–base pairs for the following reactions in aqueous solution: (i) HS−(aq) + HF ⇌ F−(aq) + H2S(aq) (ii) HPO4^2− + SO4^2− ⇌ PO4^3− + HSO4− (iii) NH3 + HCO3− ⇌ NH4+ + CO3^2−v

Solution

(i) HF donates H+ to HS−: HF is the acid, F− its conjugate base; HS− is the base, H2S its conjugate acid. (ii) HPO4^2− donates H+ to SO4^2− producing PO4^3− and HSO4−: conjugate pairs are HPO4^2−/PO4^3− and HSO4−/SO4^2−. (iii) HCO3− donates H+ to NH3 forming NH4+ and CO3^2−: conjugate pairs are NH4+/NH3 and HCO3−/CO3^2−.

Answer:

(i) HF/F− and H2S/HS− (ii) HPO4^2−/PO4^3− and HSO4−/SO4^2− (iii) NH4+/NH3 and HCO3−/CO3^2−

Q.4Account for the acidic nature of HClO4 in terms of Bronsted–Lowry theory. Identify its conjugate base.v

Solution

HClO4 → H+ + ClO4−. The conjugate base ClO4− is stabilized by delocalization of the negative charge over four oxygens (resonance), making proton donation essentially irreversible (strong acid).

Answer:

HClO4 is a strong Bronsted–Lowry acid because it readily donates H+ to solvents (e.g. water), producing H3O+. Its conjugate base is ClO4− (perchlorate), which is highly stabilized by resonance.

Q.5When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetramminecopper(II) complex, [Cu(H2O)4]2+ + 4NH3 ⇌ [Cu(NH3)4]2+ + 4H2O. Between H2O and NH3 which is the stronger Lewis base?v

Solution

NH3 donates its lone pair on nitrogen more readily than water donates a lone pair on oxygen (N is less electronegative and the lone pair is more available). Thus NH3 replaces H2O in the Cu2+ coordination sphere to give the deep-blue complex.

Answer:

NH3 is the stronger Lewis base.

Q.6The concentration of hydroxide ion in a water sample is found to be 2.5 × 10^-6 M. Identify the nature of the solution.v

Solution

pOH = −log[OH−] = −log(2.5×10^-6) ≈ 5.60. pH = 14 − pOH ≈ 14 − 5.60 = 8.40 > 7, so the solution is basic.

Answer:

Basic (pH ≈ 8.40).

Q.7A lab assistant prepared a solution by adding a calculated quantity of HCl gas at 25 °C to get a solution with [H3O+] = 4 × 10^-5 M. Is the solution neutral, acidic, or basic?v

Solution

pH = −log(4×10^-5) ≈ 4.40 (< 7), so the solution is acidic.

Answer:

Acidic.

Q.8Calculate the pH of 0.04 M HNO3 solution.v

Solution

HNO3 is a strong acid: [H+] = 0.04 M. pH = −log(0.04) = −log(4×10^-2) ≈ 1.398 ≈ 1.40.

Answer:

pH ≈ 1.40

Q.9Define solubility product.v

Solution

Ksp is the equilibrium constant for the dissolution of a slightly soluble ionic solid; it equals the product of the equilibrium concentrations of the constituent ions raised to the power of their stoichiometric coefficients.

Answer:

For a sparingly soluble salt M_aA_b(s) ⇌ a M^{n+} + b A^{m−}, the solubility product Ksp = [M^{n+}]^a [A^{m−}]^b (at saturation).

Q.10Define ionic product of water. Give its value at room temperature.v

Solution

Kw is the equilibrium constant for H2O ⇌ H+ + OH− and equals 1.0×10^-14 at 25 °C (room temperature).

Answer:

Ionic product of water Kw = [H+][OH−]; at 25 °C Kw = 1.0 × 10^-14.

Q.11Explain common ion effect with an example.v

Solution

According to Le Chatelier's principle, adding a common ion shifts the equilibrium to reduce the change. In ionic equilibria this lowers the degree of ionization or solubility of the compound that shares that ion.

Answer:

Addition of an ion common to an equilibrium suppresses the ionization or solubility of the weak electrolyte. Example: Adding NaCl to a saturated AgCl solution (which provides Cl−) decreases AgCl solubility because AgCl ⇌ Ag+ + Cl− shifts left.

Q.12Derive an expression for Ostwald's dilution law for a weak acid.v

Solution

HA ⇌ H+ + A−. Initial: c, 0, 0. At equilibrium: c(1−α), cα, cα. Ka = [H+][A−]/[HA] = (cα)(cα)/[c(1−α)] = (α^2 c)/(1−α). For very weak acids α ≪ 1, Ka ≈ α^2 c, hence α ≈ \sqrt{Ka/c}. This relation is Ostwald's dilution law.

Answer:

For a weak acid HA at concentration c with degree of dissociation α: Ka = \dfrac{α^2 c}{1−α}.

Q.13Define pH.v

Solution

pH is a measure of acidity; lower pH means higher [H+]. At 25 °C, neutral water has pH 7 (since [H+] = 1.0×10^-7 M).

Answer:

pH = −log10[H+], where [H+] is the hydrogen ion concentration in mol L^-1.

Q.14Calculate the pH of 1.5×10^-3 M solution of Ba(OH)2.v

Solution

Ba(OH)2 dissociates completely to give 2 OH− per formula unit: [OH−] = 2×1.5×10^-3 = 3.0×10^-3 M. pOH = −log(3.0×10^-3) ≈ 2.523. pH = 14 − 2.523 = 11.477 ≈ 11.48.

Answer:

pH ≈ 11.48

Q.1550 mL of 0.05 M HNO3 is mixed with 50 mL of 0.025 M KOH. Calculate the pH of the resultant solution.v

Solution

Moles H+ = 0.050 L × 0.05 M = 0.00250 mol. Moles OH− = 0.050 L × 0.025 M = 0.00125 mol. Excess H+ = 0.00125 mol in total volume 0.100 L ⇒ [H+] = 0.00125/0.100 = 0.0125 M. pH = −log(0.0125) ≈ 1.903 ≈ 1.90.

Answer:

pH ≈ 1.90

Q.16The Ka value for HCN is 10^-9. What is the pH of 0.4 M HCN solution?v

Solution

For HA ⇌ H+ + A−, [H+] ≈ \sqrt{Ka·c} = \sqrt{10^-9 × 0.4} = 2.00×10^-5 M. pH = −log(2.00×10^-5) ≈ 4.699 ≈ 4.70.

Answer:

pH ≈ 4.70

Q.17Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate. Given that Ka (acetic acid) = Kb (NH3) = 1.8×10^-5 and Kw = 1.0×10^-14.v

Solution

For ammonium acetate with equal acid and base strengths, Ka(acid)=Kb(base). Degree of hydrolysis h = \sqrt{K_w/(K_a·c)} = \sqrt{1.0×10^-14/(1.8×10^-5×0.1)} = \sqrt{5.5556×10^-9} ≈ 7.45×10^-5. Both ions hydrolyse to equal extents producing equal amounts of H+ and OH−, so they cancel and the solution remains neutral: pH = 7 (at 25 °C).

Answer:

Extent of hydrolysis h ≈ 7.45×10^-5; pH = 7.00 (neutral).

Q.18Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak basev

Solution

Consider a salt formed from a strong acid and a weak base; e.g. BH+ X– (typical: NH4+Cl–). The cation BH+ hydrolyses: \[\mathrm{BH^+ + H_2O \rightleftharpoons B + H_3O^+}\] Define Kh (hydrolysis constant): \[K_h=\frac{[B][H_3O^+]}{[BH^+]}\] But this is exactly the acid dissociation of BH+ (Ka of BH+): \[\mathrm{BH^+ \rightleftharpoons B + H^+}\quad\Rightarrow\quad K_a=\frac{[B][H^+]}{[BH^+]}=K_h.\] If B has base constant Kb, then from the relation Ka·Kb = Kw, we get \[K_h=K_a=\frac{K_w}{K_b}.\] For a salt of concentration c (initial [BH+] = c), let x = amount hydrolysed. At equilibrium [H3O+] = x, [B] = x, [BH+] = c − x. Then \[K_h=\frac{x^2}{c-x}.\] If degree of hydrolysis h = x/c and x << c, approximate c − x ≈ c, so \[x\approx\sqrt{K_h c} \quad\text{and}\quad h=\frac{x}{c}\approx\sqrt{\frac{K_h}{c}}.\] Substituting Kh = Kw/Kb gives \[h\approx\sqrt{\frac{K_w}{K_b\,c}}.\]

Answer:

Hydrolysis constant Kh = Ka = Kw/Kb. Degree of hydrolysis h ≈ sqrt(Ka/c) = sqrt(Kw/(Kb c)).

Q.19Solubility product of Ag2CrO4 is 1.0×10^-12. What is the solubility of Ag2CrO4 in 0.01 M AgNO3 solution?v

Solution

Dissolution: \(\mathrm{Ag_2CrO_4\rightleftharpoons 2Ag^+ + CrO_4^{2-}}\). Let solubility be s in the presence of [Ag+]_initial = 0.01 M. Then [Ag+] = 0.01 + 2s ≈ 0.01, [CrO4^{2-}] = s. Ksp = [Ag+]^2[CrO4^{2-}] = (0.01)^2·s = 1.0\times10^{-4}s. So s = Ksp / 1.0\times10^{-4} = 1.0\times10^{-12}/1.0\times10^{-4} = 1.0\times10^{-8} M.

Answer:

1.0×10^-8 M

Q.20Write the expression for the solubility product of Ca3(PO4)2v

Solution

Dissociation: \(\mathrm{Ca_3(PO_4)_2\rightleftharpoons 3Ca^{2+}+2PO_4^{3-}}\). Therefore \[K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2.\]

Answer:

Ksp = [Ca2+]^3[PO4^3-]^2

Q.21A saturated solution, prepared by dissolving CaF2(s) in water, has [Ca2+] = 3.3×10^-4 M. What is the Ksp of CaF2 ?v

Solution

Dissolution: \(\mathrm{CaF_2\rightleftharpoons Ca^{2+}+2F^-}\). Given [Ca^{2+}] = 3.3×10^{-4} M, so [F^-] = 2[Ca^{2+}] = 6.6×10^{-4} M. Ksp = [Ca^{2+}][F^-]^2 = (3.3×10^{-4})·(6.6×10^{-4})^2 = 3.3×10^{-4}·(4.356×10^{-7}) ≈ 1.44×10^{-10}.

Answer:

1.44×10^-10

Q.22Ksp of AgCl is 1.8×10^-10. Calculate molar solubility in 1 M AgNO3v

Solution

Dissolution: \(\mathrm{AgCl\rightleftharpoons Ag^+ + Cl^-}\). With [Ag^+] = 1.00 M (common ion), Ksp = [Ag^+][Cl^-] = 1.00·s, so s = Ksp = 1.8×10^{-10} M.

Answer:

1.8×10^-10 M

Q.23A particular saturated solution of silver chromate Ag2CrO4 has [Ag+] = 5.0×10^-5 M and [CrO4^2-] = 4.4×10^-4 M. What is the value of Ksp for Ag2CrO4?v

Solution

For \(\mathrm{Ag_2CrO_4}\): Ksp = [Ag^+]^2[CrO_4^{2-}]. Ksp = (5.0×10^{-5})^2·(4.4×10^{-4}) = 2.5×10^{-9}·4.4×10^{-4} = 1.10×10^{-12}.

Answer:

1.1×10^-12

Q.24Write the expression for the solubility product of Hg2Cl2.v

Solution

Dissociation (mercurous chloride): \(\mathrm{Hg_2Cl_2\rightleftharpoons Hg_2^{2+} + 2Cl^-}\). Hence Ksp = [Hg_2^{2+}][Cl^-]^2.

Answer:

Ksp = [Hg2^{2+}][Cl^-]^2

Q.25Ksp of Ag2CrO4 is 1.1×10^-12. What is solubility of Ag2CrO4 in 0.1 M K2CrO4?v

Solution

Dissolution: \(\mathrm{Ag_2CrO_4\rightleftharpoons 2Ag^+ + CrO_4^{2-}}\). Let solubility be s. In 0.10 M CrO4^{2-} (common ion), [CrO4^{2-}] ≈ 0.10, [Ag^+] = 2s. Ksp = (2s)^2(0.10) = 4s^2·0.10 = 0.4 s^2. So s^2 = Ksp/0.4 = 1.1×10^{-12}/0.4 = 2.75×10^{-12} ⇒ s = √(2.75×10^{-12}) ≈ 1.66×10^{-6} M.

Answer:

1.66×10^-6 M

Q.26Will a precipitate be formed when 0.150 L of 0.1 M Pb(NO3)2 and 0.100 L of 0.2 M NaCl are mixed? Ksp(PbCl2)=1.2×10^-5v

Solution

Moles Pb^{2+} = 0.150·0.1 = 0.015 mol; moles Cl^- = 0.100·0.2 = 0.020 mol. Total volume = 0.250 L. [Pb^{2+}] = 0.015/0.25 = 0.06 M; [Cl^-] = 0.020/0.25 = 0.08 M. Q = [Pb^{2+}][Cl^-]^2 = 0.06·(0.08)^2 = 0.06·0.0064 = 3.84×10^{-4}. Since Q(=3.84×10^{-4}) > Ksp(=1.2×10^{-5}), precipitation of PbCl2 will occur.

Answer:

Yes, a precipitate will form (Qsp = 3.84×10^-4 > Ksp).

Q.27Ksp of Al(OH)3 is 1.0×10^-33. At what pH does 1.0×10^-3 M Al3+ precipitate on the addition of buffer of NH4Cl and NH4OH solution?v

Solution

Precipitation occurs when [Al^{3+}][OH^-]^3 ≥ Ksp. Given [Al^{3+}] = 1.0×10^{-3} M and Ksp = 1.0×10^{-33}. Solve for [OH^-]_{crit}: \[[OH^-] = \left(\frac{K_{sp}}{[Al^{3+}]}\right)^{1/3} = \left(\frac{1.0\times10^{-33}}{1.0\times10^{-3}}\right)^{1/3} = (1.0\times10^{-30})^{1/3} = 1.0\times10^{-10}\,M.\] pOH = 10.00 ⇒ pH = 14.00 − 10.00 = 4.00. Thus Al(OH)3 begins to precipitate when the pH exceeds about 4.00.

Answer:

pH = 4.00 (precipitation occurs when pH > 4.00)

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