Every Book Back multiple-choice question from Ionic Equilibrium (12th Standard Chemistry, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
Concentration of the \(Ag^+\) ions in a saturated solution of \(Ag_2C_2O_4\) is \(2.24 \times 10^{-4}\text{ mol L}^{-1}\) solubility product of \(Ag_2C_2O_4\) is (NEET – 2017)
- A. \(2.42 \times 10^{-8}\text{ mol}^3\text{L}^{-3}\)
- B. \(2.66 \times 10^{-12}\text{ mol}^3\text{L}^{-3}\)
- C. \(4.5 \times 10^{-11}\text{ mol}^3\text{L}^{-3}\)
- D. \(5.619 \times 10^{-12}\text{ mol}^3\text{L}^{-3}\)Correct
Explanation. The silver ion concentration represents \(2s\) in the solubility expression. Solving for \(s\) and substituting into the \(K_{sp}\) formula \(4s^3\) gives the final product value.
Q2
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations. (NEET – 2018)
i. 60 mL \(\frac{M}{10}\) HCl + 40mL \(\frac{M}{10}\) NaOH
ii. 55 mL \(\frac{M}{10}\) HCl + 45 mL \(\frac{M}{10}\) NaOH
iii. 75 mL \(\frac{M}{5}\) HCl + 25mL \(\frac{M}{5}\) NaOH
iv. 100 mL \(\frac{M}{10}\) HCl + 100 mL \(\frac{M}{10}\) NaOH
pH of which one of them will be equal to 1?
- A. iv
- B. i
- C. ii
- D. iiiCorrect
Explanation. A pH of 1 corresponds to a hydrogen ion concentration of \(0.1M\). Calculating the remaining acid concentration after neutralization for the third mixture yields exactly \(0.1M\).
Q3
The solubility of \(BaSO_4\) in water is \(2.42 \times 10^{-3}\text{ gL}^{-1}\) at 298K. The value of its solubility product \(K_{sp}\) will be (NEET -2018). (Given molar mass of \(BaSO_4 = 233\text{g mol}^{-1}\))
- A. \(1.08 \times 10^{-14}\text{ mol}^2\text{ L}^{-2}\)
- B. \(1.08 \times 10^{-12}\text{ mol}^2\text{ L}^{-2}\)
- C. \(1.08 \times 10^{-10}\text{ mol}^2\text{ L}^{-2}\)Correct
- D. \(1.08 \times 10^{-8}\text{ mol}^2\text{ L}^{-2}\)
Explanation. First, convert the solubility from grams per liter to moles per liter using the molar mass. Then, since the salt dissociates into two ions, square this molar solubility to find the solubility product.
Q4
pH of a saturated solution of \(Ca(OH)_2\) is 9. The Solubility product (\(K_{sp}\)) of \(Ca(OH)_2\) is
- A. \(0.5 \times 10^{-15}\)Correct
- B. \(0.25 \times 10^{-10}\)
- C. \(0.125 \times 10^{-15}\)
- D. \(0.5 \times 10^{-10}\)
Explanation. Starting from pH, determine pOH and the hydroxide ion concentration. Since the calcium ion concentration is half that of the hydroxide, use the equilibrium expression to calculate the solubility product.
Q5
Conjugate base for Bronsted acids \(H_2O\) and HF are
- A. \(OH^-\) and \(H_2F^+\), respectively
- B. \(H_3O^+\) and \(F^-\), respectively
- C. \(OH^-\) and \(F^-\), respectivelyCorrect
- D. \(H_3O^+\) and \(H_2F^+\), respectively
Explanation. A conjugate base is formed when a Bronsted acid donates a proton. Removing one hydrogen ion from water and hydrogen fluoride results in the hydroxide and fluoride ions.
Q6
Which will make basic buffer?
- A. 50 mL of 0.1M NaOH + 25mL of 0.1M \(CH_3COOH\)
- B. 100 mL of 0.1M \(CH_3COOH\) + 100 mL of 0.1M \(NH_4OH\)
- C. 100 mL of 0.1M HCl + 200 mL of 0.1M \(NH_4OH\)Correct
- D. 100 mL of 0.1M HCl + 100 mL of 0.1M NaOH
Explanation. A basic buffer requires a mixture of a weak base and its salt with a strong acid. Mixing excess ammonium hydroxide with hydrochloric acid creates this specific system.
Q7
Which of the following fluro compounds is most likely to behave as a Lewis base? (NEET – 2016)
- A. \(BF_3\)
- B. \(PF_3\)Correct
- C. \(CF_4\)
- D. \(SiF_4\)
Explanation. A Lewis base must possess an unshared pair of electrons to donate. Phosphorus trifluoride satisfies this requirement, as the central phosphorus atom has a lone pair available.
Q8
Which of these is not likely to act as Lewis base?
- A. \(BF_3\)Correct
- B. \(PF_3\)
- C. CO
- D. \(F^-\)
Explanation. Boron trifluoride is an electron-deficient molecule with an incomplete octet. Therefore, it acts as a Lewis acid by accepting electrons rather than behaving as a Lewis base.
Q9
The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively
- A. acidic, acidic, basic
- B. basic, acidic, basicCorrect
- C. basic, neutral, basic
- D. none of these
Explanation. The pH of salt solutions depends on the strength of their parent acids and bases. Salts of strong bases and weak acids are basic, while salts of strong acids and weak bases are acidic.
Q10
The percentage of pyridine (\(C_5H_5N\)) that forms pyridinium ion (\(C_5H_5NH^+\)) in a 0.10M aqueous pyridine solution (\(K_b\) for \(C_5H_5N = 1.7 \times 10^{-9}\)) is
- A. 0.006%
- B. 0.013%Correct
- C. 0.77%
- D. 1.6%
Explanation. Calculate the degree of dissociation using the square root of the ratio of the base dissociation constant to the concentration. Multiplying this value by one hundred provides the dissociation percentage.
Q11
Equal volumes of three acid solutions of pH 1, 2 and 3 are mixed in a vessel. What will be the \(H^+\) ion concentration in the mixture?
- A. \(3.7 \times 10^{-2}\)Correct
- B. \(10^{-6}\)
- C. 0.111
- D. none of these
Explanation. Find the hydrogen ion concentration for each pH value. The final concentration in the mixture is the average of these three concentrations, assuming equal volumes were used for mixing.
Q12
The solubility of AgCl (s) with solubility product \(1.6 \times 10^{-10}\) in 0.1M NaCl solution would be
- A. \(1.26 \times 10^{-5} M\)
- B. \(1.6 \times 10^{-9} M\)Correct
- C. \(1.6 \times 10^{-11} M\)
- D. Zero
Explanation. The presence of a common chloride ion from sodium chloride significantly reduces the solubility of silver chloride. Solve the solubility product expression using the known chloride concentration from the salt.
Q13
If the solubility product of lead iodide is \(3.2 \times 10^{-8}\), its solubility will be
- A. \(2 \times 10^{-3} M\)Correct
- B. \(4 \times 10^{-4} M\)
- C. \(1.6 \times 10^{-5} M\)
- D. \(1.8 \times 10^{-5} M\)
Explanation. For lead iodide (\(PbI_2\)), the solubility product \(K_{sp}\) is related to solubility \(s\) by \(K_{sp} = 4s^3\). Solving \(3.2 \times 10^{-8} = 4s^3\) gives \(s^3 = 8 \times 10^{-9}\), so \(s = 2 \times 10^{-3} M\).
Q14
MY and \(NY_3\), are insoluble salts and have the same \(K_{sp}\) values of \(6.2 \times 10^{-13}\) at room temperature. Which statement would be true with regard to MY and \(NY_3\)?
- A. The salts MY and \(NY_3\) are more soluble in 0.5M KY than in pure water
- B. The addition of the salt of KY to the suspension of MY and \(NY_3\) will have no effect on their solubilities
- C. The molar solubilities of MY and \(NY_3\) in water are identical
- D. The molar solubility of MY in water is less than that of \(NY_3\)Correct
Explanation. Solubility \(s\) is calculated as \(\sqrt{K_{sp}}\) for MY and \((K_{sp}/27)^{1/4}\) for \(NY_3\). For the same \(K_{sp}\) value, the fourth root calculation for \(NY_3\) result in a higher value than the square root for MY.
Q15
What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?
- A. 2.0
- B. 3
- C. 7.0
- D. 12.65Correct
Explanation. The mixture has excess base. The resulting \([OH^-]\) concentration is \((0.1 - 0.01) / 2 = 0.045 M\). Calculating \(pOH = -\log(0.045) \approx 1.35\), then \(pH = 14 - 1.35 = 12.65\).
Q16
The dissociation constant of a weak acid is \(1 \times 10^{-3}\). In order to prepare a buffer solution with a pH = 4, the \([Acid]/[Salt]\) ratio should be
- A. 4:3
- B. 3:4
- C. 10:1
- D. 1:10Correct
Explanation. Using Henderson-Hasselbalch equation: \(pH = pK_a + \log([Salt]/[Acid])\). Given \(pH=4\) and \(pK_a=3\), then \(1 = \log([Salt]/[Acid])\), so \([Salt]/[Acid] = 10\), and \([Acid]/[Salt] = 1/10\).
Q17
The pH of \(10^{-5}\) M KOH solution will be
- A. 9Correct
- B. 5
- C. 19
- D. none of these
Explanation. For \(10^{-5} M\) KOH, the \([OH^-]\) is \(10^{-5}\). Thus \(pOH = 5\). Since \(pH + pOH = 14\), the \(pH = 14 - 5 = 9\).
Q18
\(H_2PO_4^-\) is the conjugate base of
- A. \(PO_4^{3-}\)
- B. \(P_2O_5\)
- C. \(H_3PO_4\)Correct
- D. \(HPO_4^{2-}\)
Explanation. A conjugate base is formed when an acid donates a proton (\(H^+\)). \(H_3PO_4\) losing one proton becomes its conjugate base, \(H_2PO_4^-\).
Q19
Which of the following can act as Lowry – Bronsted acid as well as base?
- A. HCl
- B. \(SO_4^{2-}\)
- C. \(HPO_4^{2-}\)Correct
- D. \(Br^-\)
Explanation. An amphoteric species like \(HPO_4^{2-}\) can act as an acid by donating its remaining proton to form \(PO_4^{3-}\) or act as a base by accepting a proton to form \(H_2PO_4^-\).
Q20
The pH of an aqueous solution is Zero. The solution is
- A. slightly acidic
- B. strongly acidicCorrect
- C. neutral
- D. basic
Explanation. A pH of 0 corresponds to a hydronium ion concentration of \(10^0 = 1 M\), which indicates a highly concentrated acid solution, thus it is strongly acidic.
Q21
The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by
- A. \([H^+] = K_a \frac{[acid]}{[salt]}\)Correct
- B. \([H^+] = K_a [salt]\)
- C. \([H^+] = K_a [acid]\)
- D. \([H^+] = K_a \frac{[salt]}{[acid]}\)
Explanation. Based on the buffer equation derived from the acid dissociation constant expression \(K_a = \frac{[H^+][salt]}{[acid]}\), rearranging for \([H^+]\) gives \([H^+] = K_a \frac{[acid]}{[salt]}\).
Q22
Which of the following relation is correct for degree of hydrolysis of ammonium acetate?
- A. \(h = \sqrt{\frac{K_h}{C}}\)
- B. \(h = \sqrt{\frac{K_a}{K_b}}\)
- C. \(h = \sqrt{\frac{K_w}{K_a \cdot K_b}}\)Correct
- D. \(h = \sqrt{\frac{K_a \cdot K_b}{K_w}}\)
Explanation. Ammonium acetate is a salt of a weak acid and a weak base. Its degree of hydrolysis \(h\) is independent of concentration and relates to \(K_w, K_a,\) and \(K_b\) by \(h = \sqrt{\frac{K_w}{K_a \cdot K_b}}\).
Q23
Dissociation constant of \(NH_4OH\) is \(1.8 \times 10^{-5}\) the hydrolysis constant of \(NH_4Cl\) would be
- A. \(1.8 \times 10^{-19}\)
- B. \(5.55 \times 10^{-10}\)Correct
- C. \(5.55 \times 10^{-5}\)
- D. \(1.80 \times 10^{-5}\)
Explanation. For a salt of a strong acid and weak base like \(NH_4Cl\), the hydrolysis constant \(K_h = K_w / K_b\). Using \(10^{-14} / (1.8 \times 10^{-5})\) results in \(5.55 \times 10^{-10}\).
