Q.Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10⁻⁸ ).

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The mass of $CaC_2O_4$ (in $g$) to be dissolved in distilled water to make 1.0 L of saturated solution is ($K_{SP}$ of $CaC_2O_4 = 2.5 \times 10^{-9}\ mol^2L^{-2}$, molar mass of $CaC_2O_4 = 128\ g\,mol^{-1}$) (A) 0.0064 (B) 0.0032 (C) 0.0128 (D) 0.0640

   ›Reveal solutionSolution

   A standard solubility-product calculation for a 1:1 salt; the mass dissolved in 1.0 L is 0.0064 g.

   Concept and Intuition

   For a sparingly soluble salt $MX$ that dissociates 1:1, the solubility product $K_{sp}=[M^{2+}][X^{2-}]=s^2$, where $s$ is the molar solubility. Once $s$ is known, converting to mass just needs the molar mass and the volume of solution.

   Step-by-Step Solution

   1. Dissociation: $CaC_2O_4(s) \rightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)$; both ions appear with coefficient 1, so $K_{sp}=s^2$.
   2. $s=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-9}}=\sqrt{25\times10^{-10}}=5\times10^{-5}$ mol/L.
   3. Moles in 1.0 L $= 5\times10^{-5}\times1.0 = 5\times10^{-5}$ mol. …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The minimum volume of water (in $L$) needed to dissolve 1.5 $g$ of $CaSO_4$ (molar mass = $136\ g\,mol^{-1}$) at 298 $K$ is ($K_{sp}(CaSO_4)$ at 298 $K$ = $9 \times 10^{-6}$) (A) $6.37$ (B) $7.37$ (C) $3.67$ (D) $3.73$

   ›Reveal solutionSolution

   This tests using $K_{sp}$ to find molar solubility, then using solubility as a concentration ceiling to find the minimum volume needed to dissolve a given mass without exceeding saturation. The minimum volume is $3.67\ L$.

   Concept and Intuition

   "Minimum volume needed to dissolve" a given mass of a sparingly soluble salt means: what's the smallest volume in which this mass, fully dissociated, still doesn't exceed the saturation concentration set by $K_{sp}$? At exactly this volume, the solution is saturated (just at the point of precipitation) — any less volume and the salt could not stay fully dissolved.

   Step-by-Step Solution

   1. Moles of $CaSO_4$: $n = \dfrac{1.5\ g}{136\ g\,mol^{-1}} = 0.011029\ mol$.
   2. For a 1:1 dissociation $CaSO_4 \rightleftharpoons Ca^{2+} + SO_4^{2-}$, if molar solubility is $s$, then $K_{sp} = [Ca^{2+}][SO_4^{2-}] = s^2$.
   3. Solve for $s$: $s = \sqrt{K_{sp}} = \sqrt{9\times10^{-6}} = 3\times10^{-3}\ mol\,L^{-1}$. …

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3. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The molar solubility of $\mathrm{PbI_2}$ in 0.2 M $\mathrm{Pb(NO_3)_2}$ solution in terms of $K_{sp}$ (solubility product) is (A) $\left(\dfrac{K_{sp}}{0.2}\right)^{1/2}$ (B) $\left(\dfrac{K_{sp}}{0.4}\right)^{1/4}$ (C) $\left(\dfrac{K_{sp}}{0.8}\right)^{1/2}$ (D) $\left(\dfrac{K_{sp}}{0.8}\right)^{1/3}$

   ›Reveal solutionSolution

   This tests the common-ion effect on solubility product calculations; molar solubility works out to $\left(\dfrac{K_{sp}}{0.8}\right)^{1/2}$.

   Concept and Intuition

   When a sparingly soluble salt like $\mathrm{PbI_2}$ is dissolved into a solution that already contains one of its constituent ions (here, $\mathrm{Pb^{2+}}$ from $\mathrm{Pb(NO_3)_2}$), Le Chatelier's principle tells us the equilibrium shifts to suppress dissolution — this is the common-ion effect. Since the pre-existing $\mathrm{Pb^{2+}}$ concentration (0.2 M) is much larger than the tiny extra amount contributed by the sparingly soluble $\mathrm{PbI_2}$, we can treat $[\mathrm{Pb^{2+}}]$ as essentially fixed at 0.2 M.

   Step-by-Step Solution

   1. Dissolution equilibrium: $\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)}$.
   2. Let molar solubility of $\mathrm{PbI_2}$ in this solution be $s$. This contributes $s$ mol/L of $\mathrm{Pb^{2+}}$ and $2s$ mol/L of $\mathrm{I^-}$.
   3. Since 0.2 M $\mathrm{Pb^{2+}}$ already exists in solution and $s$ is very small compared to 0.2, total $[\mathrm{Pb^{2+}}] \approx 0.2 + s \approx 0.2$ M.
   4. $[\mathrm{I^-}] = 2s$. …

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4. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.At T(K) $K_{sp}$ of two ionic salts $\mathrm{MX_2}$ and MX is $5\times10^{-13}$ and $1.6\times10^{-11}$ respectively. The ratio of molar solubility of $\mathrm{MX_2}$ and MX is (A) 12.5 (B) 1.25 (C) 6.25 (D) 7.50

   ›Reveal solutionSolution

   Molar solubility must be derived from each salt's own dissociation stoichiometry ($K_{sp}=4s^3$ for $MX_2$, $K_{sp}=s^2$ for $MX$) before comparing — doing that gives a solubility ratio of 12.5.

   Concept and Intuition

   $K_{sp}$ values of two salts with different dissociation patterns cannot be compared directly to get a solubility ratio — the exponent and stoichiometric factor in the $K_{sp}$ expression depend on how many ions of each type are released. $MX_2 \rightleftharpoons M^{2+} + 2X^-$ gives $K_{sp}=[M^{2+}][X^-]^2 = s(2s)^2 = 4s^3$, while $MX \rightleftharpoons M^+ + X^-$ gives $K_{sp}=s\cdot s = s^2$.

   Step-by-Step Solution

   1. For $MX_2$: $K_{sp}=4s_1^3 = 5\times10^{-13} \Rightarrow s_1^3 = 1.25\times10^{-13} = 125\times10^{-15}$.
   2. $s_1 = \sqrt[3]{125\times10^{-15}} = 5\times10^{-5}\ \text{mol/L}$.
   3. For $MX$: $K_{sp}=s_2^2 = 1.6\times10^{-11} \Rightarrow s_2 = \sqrt{16\times10^{-12}} = 4\times10^{-6}\ \text{mol/L}$. …

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5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The solubility of barium phosphate of molar mass 'M' g mol$^{-1}$ in water is $x$ g per 100 mL at 298 K. Its solubility product is $1.08 \times \left(\dfrac{x}{M}\right)^a \times (10)^b$. The values of a and b respectively are (A) 7, 5 (B) 5, 7 (C) 5, 5 (D) 7, 7

   ›Reveal solutionSolution

   Writing the $K_{sp}$ expression for $Ba_3(PO_4)_2$ in terms of molar solubility and converting units carefully gives exponents $a=5$, $b=7$.

   Concept and Intuition

   $K_{sp}$ depends on the stoichiometric powers of the ions in the dissolution equilibrium. The trick in this problem is purely unit conversion: solubility is given as grams per 100 mL, but $K_{sp}$ needs molar concentration (mol/L).

   Step-by-Step Solution

   1. Dissolution: $Ba_3(PO_4)_2(s) \rightleftharpoons 3Ba^{2+} + 2PO_4^{3-}$.
   2. If molar solubility is $s$ mol/L: $[Ba^{2+}] = 3s$, $[PO_4^{3-}] = 2s$.
   3. $K_{sp} = (3s)^3(2s)^2 = 27s^3 \times 4s^2 = 108\,s^5$.
   4. Given solubility $= x$ g per 100 mL with molar mass $M$: moles per 100 mL $= x/M$, so moles per litre $s = \dfrac{x}{M}\times\dfrac{1000}{100} = \dfrac{10x}{M}$.
   5. Substitute: $K_{sp} = 108\left(\dfrac{10x}{M}\right)^5 = 108\times 10^5 \times \left(\dfrac{x}{M}\right)^5$. …

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6. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.At T(K), $K_{sp}$ of AgCl, AgBr and AgI is $1.8\times10^{-10}$, $5\times10^{-13}$ and $8.3\times10^{-17}$ respectively. A 1L solution contains $10^{-5}$ moles each of NaCl, NaBr and NaI. This solution was titrated with $10^{-6}$ M $\mathrm{AgNO_3}$ solution, till all the $\mathrm{Cl^-}$, $\mathrm{Br^-}$ and $\mathrm{I^-}$ are precipitated as silver salts. The order of precipitation of these respectively are (A) AgCl, AgBr, AgI (B) AgI, AgBr, AgCl (C) AgI, AgCl, AgBr (D) All the three salts are precipitated simultaneously

   ›Reveal solutionSolution

   Comparing $[Ag^+]$ required to reach each salt's $K_{sp}$ at the same anion concentration shows AgI needs the least silver ion and precipitates first, followed by AgBr, then AgCl.

   Concept and Intuition

   For a set of sparingly soluble salts sharing the same cation being slowly titrated in, and starting with equal anion concentrations, the salt that requires the smallest $[Ag^+]$ to reach its solubility-product threshold will start precipitating first. This required concentration is $[Ag^+]_{required} = K_{sp}/[\text{anion}]$ — so the salt with the smallest $K_{sp}$ (given equal anion concentrations) precipitates earliest.

   Step-by-Step Solution

   1. $[Ag^+]$ needed for AgCl to start precipitating: $\dfrac{1.8\times10^{-10}}{10^{-5}} = 1.8\times10^{-5}$ M.
   2. $[Ag^+]$ needed for AgBr: $\dfrac{5\times10^{-13}}{10^{-5}} = 5\times10^{-8}$ M.
   3. $[Ag^+]$ needed for AgI: $\dfrac{8.3\times10^{-17}}{10^{-5}} = 8.3\times10^{-12}$ M. …

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7. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.Observe the following solutions

   i) 1L of $10^{-6}$ M $\mathrm{AgNO_3}$ ii) 1L of $10^{-7}$ M $\mathrm{AgNO_3}$ iii) 1L of $10^{-9}$ M $\mathrm{AgNO_3}$

   iv) 1L of $10^{-3}$ M $\mathrm{AgNO_3}$ v) 1L of $10^{-5}$ M NaCl Which of the above two solutions when mixed will give a white precipitate, AgCl? (Given $K_{sp}$ of AgCl = $1\times10^{-10}$) (A) (i),

   (v) (B) (ii),

   (v) (C) (iv),

   (v) (D) (iii), (v)

   ›Reveal solutionSolution

   A precipitate forms only when the ionic product $[\text{Ag}^+][\text{Cl}^-]$ (after mixing and dilution) exceeds $K_{sp}$; checking each pairing shows only $10^{-3}$ M AgNO$_3$ mixed with $10^{-5}$ M NaCl exceeds $K_{sp} = 10^{-10}$.

   Concept and Intuition

   Whether a precipitate forms when two solutions are mixed is decided by comparing the "ionic product" (reaction quotient $Q_{sp}$) of the relevant ions, at the moment of mixing, to the solubility product $K_{sp}$. If $Q_{sp} > K_{sp}$, the solution is supersaturated and precipitation occurs; if $Q_{sp} \leq K_{sp}$, no precipitate forms — the solution remains unsaturated (or just saturated).

   Step-by-Step Solution

   1. When 1 L of an AgNO$_3$ solution is mixed with 1 L of the NaCl solution (v), the total volume becomes 2 L, so each original concentration is halved on mixing.
   2. Compute $Q_{sp} = [\text{Ag}^+][\text{Cl}^-]$ after mixing for each AgNO$_3$ option with (v) $10^{-5}$ M NaCl (halved to $0.5\times10^{-5}$ M):
      • (i) $10^{-6}$ M AgNO$_3$ → $[\text{Ag}^+]=0.5\times10^{-6}$: $Q_{sp} = 0.5\times10^{-6}\times0.5\times10^{-5} = 2.5\times10^{-12}$ — less than $K_{sp}=10^{-10}$, no ppt.
      • (ii) $10^{-7}$ M → $[\text{Ag}^+]=0.5\times10^{-7}$: $Q_{sp}=2.5\times10^{-13}$ — no ppt.
      • (iii) $10^{-9}$ M → even smaller $Q_{sp}$ — no ppt. …

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8. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The solubility products of NiS, ZnS, CdS and HgS are $4.7\times10^{-5}$, $1.6\times10^{-24}$, $8\times10^{-27}$ and $4\times10^{-53}$ respectively. An aqueous solution contains $Ni^{2+}$, $Zn^{2+}$, $Cd^{2+}$, and $Hg^{2+}$ of equal concentration. $H_2S$ gas was passed into this solution very slowly. The first and the last ions that precipitate as sulphides are respectively (A) $Ni^{2+}$, $Hg^{2+}$ (B) $Hg^{2+}$, $Cd^{2+}$ (C) $Zn^{2+}$, $Hg^{2+}$ (D) $Hg^{2+}$, $Ni^{2+}$

   ›Reveal solutionSolution

   When $H_2S$ is passed slowly into a mixture of metal ions at equal concentration, the sulfide with the smallest $K_{sp}$ precipitates first and the one with the largest $K_{sp}$ precipitates last. Here: $Hg^{2+}$ first, $Ni^{2+}$ last.

   Concept and Intuition

   As $H_2S$ dissolves slowly, $[S^{2-}]$ rises gradually from near zero. A given sulfide $MS$ starts to precipitate as soon as the ion product $[M^{2+}][S^{2-}]$ exceeds its $K_{sp}$. Since all the metal ions start at the same concentration, the sulfide requiring the least $[S^{2-}]$ to reach its $K_{sp}$ threshold — i.e., the one with the smallest $K_{sp}$ — crosses that threshold first. The largest-$K_{sp}$ sulfide needs the most $[S^{2-}]$ and so precipitates last.

   Step-by-Step Solution

   1. List $K_{sp}$: NiS $=4.7\times10^{-5}$, ZnS $=1.6\times10^{-24}$, CdS $=8\times10^{-27}$, HgS $=4\times10^{-53}$.
   2. Order from smallest to largest: HgS < CdS < ZnS < NiS. …

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9. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.A 1.0 L of aqueous solution contains $1 \times 10^{-8}$ M NaBr, $1 \times 10^{-8}$ M NaCl and $1 \times 10^{-8}$ M NaI. To this solution, $1 \times 10^{-10}$ M aqueous $\mathrm{AgNO_3}$ solution is added drop wise. The order of precipitation of Ag X (X = Cl, Br, I) is ($K_{SP}(\mathrm{AgCl}) = 1.8 \times 10^{-10}$; $K_{SP}(\mathrm{AgBr}) = 5 \times 10^{-13}$; $K_{SP}(\mathrm{AgI}) = 8.3 \times 10^{-17}$) (A) AgBr, AgCl, AgI (B) AgCl, AgBr, AgI (C) AgI, AgBr, AgCl (D) AgBr, AgI, AgCl

   ›Reveal solutionSolution

   With equal halide concentrations, the salt with the smallest solubility product needs the least added Ag⁺ to precipitate, giving the order AgI, then AgBr, then AgCl.

   Concept and Intuition

   When Ag⁺ is added slowly to a solution containing several halide ions at known concentrations, each silver halide begins to precipitate once the ion product $[\text{Ag}^+][\text{X}^-]$ exceeds that salt's $K_{sp}$. Since all three halides here start at the same concentration ($10^{-8}$ M), the required $[\text{Ag}^+]$ to trigger precipitation is $K_{sp}/[\text{X}^-]$ — directly proportional to $K_{sp}$. The salt with the smallest $K_{sp}$ needs the smallest $[\text{Ag}^+]$, so it precipitates first as Ag⁺ concentration slowly rises; the salt with the largest $K_{sp}$ needs the most Ag⁺ and precipitates last.

   Step-by-Step Solution

   1. Required $[\text{Ag}^+]$ for AgI to start precipitating: $\dfrac{8.3\times10^{-17}}{10^{-8}} = 8.3\times10^{-9}$ M.
   2. Required $[\text{Ag}^+]$ for AgBr: $\dfrac{5\times10^{-13}}{10^{-8}} = 5\times10^{-5}$ M.
   3. Required $[\text{Ag}^+]$ for AgCl: $\dfrac{1.8\times10^{-10}}{10^{-8}} = 1.8\times10^{-2}$ M. …

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10. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.The solubility of $AgBr_{(s)}$, having solubility product $5 \times 10^{-10}$ in $0.2\ M\ NaBr$ solution, equals (A) $5 \times 10^{-10}\ M$ (B) $25 \times 10^{-10}\ M$ (C) $0.5\ M$ (D) $0.002\ M$

    ›Reveal solutionSolution

    The common-ion effect from 0.2 M NaBr fixes [Br−]≈0.2 M, so AgBr's solubility (=[Ag+]) works out to 2.5×10⁻⁹ M = 25×10⁻¹⁰ M. Answer (B).

    Concept and Intuition

    When a sparingly soluble salt dissolves in a solution already containing one of its constituent ions (here $Br^-$ from NaBr), the common-ion effect suppresses the salt's own dissociation, sharply reducing its solubility below the value in pure water. Because NaBr fully dissociates and is present at a concentration vastly larger than AgBr's own solubility, $[Br^-]$ can be taken as essentially fixed at the NaBr concentration.

    Step-by-Step Solution

    1. $AgBr_{(s)}\rightleftharpoons Ag^+_{(aq)}+Br^-_{(aq)}$, with $K_{sp}=[Ag^+][Br^-]=5\times10^{-10}$.
    2. In 0.2 M NaBr, $[Br^-]\approx0.2$ M (the tiny extra Br⁻ from AgBr's own dissolution is negligible next to 0.2 M).
    3. Solubility of AgBr $=[Ag^+]$ (each formula unit that dissolves releases one $Ag^+$). …

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11. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.If the concentration of $Ag^+$ ions in the saturated solution of $\mathrm{Ag_2CO_3}$ is $1.20 \times 10^{-4}\ mol.L^{-1}$, then find the solubility product of $\mathrm{Ag_2CO_3}$. (A) $5.30 \times 10^{-12}$ (B) $4.50 \times 10^{-11}$ (C) $2.66 \times 10^{-12}$ (D) $6.90 \times 10^{-12}$

    ›Reveal solutionSolution

    Using the molar solubility of $Ag_2CO_3$ ($s=1.2\times10^{-4}$) with the dissociation stoichiometry $Ag_2CO_3\to 2Ag^++CO_3^{2-}$ gives $K_{sp}=4s^3\approx6.9\times10^{-12}$.

    Concept and Intuition

    $Ag_2CO_3$ dissociates as $Ag_2CO_3 \rightleftharpoons 2Ag^+ + CO_3^{2-}$. If $s$ is the molar solubility of the compound, each formula unit that dissolves releases 2 silver ions and 1 carbonate ion, so $[Ag^+]=2s$ and $[CO_3^{2-}]=s$ — the factor of 2 (squared, since it enters $K_{sp}$ as $[Ag^+]^2$) is the key stoichiometric detail this problem tests.

    Step-by-Step Solution

    1. Dissociation: $Ag_2CO_3 \to 2Ag^+ + CO_3^{2-}$.
    2. Let molar solubility $=s=1.20\times10^{-4}\ mol/L$.
    3. $[Ag^+]=2s = 2.4\times10^{-4}$, $[CO_3^{2-}]=s=1.2\times10^{-4}$. …

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12. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.If the molar solubility (in $mol.L^{-1}$) of a sparingly soluble salt $AB_4$ is 'S', and the corresponding solubility product is '$K_{sp}$', then S in terms of $K_{sp}$ is given by the relation ______ (A) $S = \left(\dfrac{K_{sp}}{128}\right)^{1/4}$ (B) $S = \left(\dfrac{K_{sp}}{256}\right)^{1/5}$ (C) $S = (256\, K_{sp})^{1/5}$ (D) $S = (128\, K_{sp})^{1/4}$

    ›Reveal solutionSolution

    Writing the dissociation of $AB_4$ in terms of molar solubility $S$ and substituting into the solubility-product expression gives $S=(K_{sp}/256)^{1/5}$.

    Concept and Intuition

    For a sparingly soluble salt $AB_n$ that dissociates into one $A^{n+}$ ion and $n$ $B^-$ ions, the solubility product $K_{sp}$ is the product of ion concentrations, each raised to its stoichiometric coefficient. Since each mole of salt that dissolves produces 1 mole of $A^{n+}$ but $n$ moles of $B^-$, the $B^-$ concentration is $n$ times the molar solubility, and this multiplicative factor gets raised to the $n^{th}$ power in the $K_{sp}$ expression — producing a large numerical coefficient.

    Step-by-Step Solution

    1. Dissociation: $AB_4 \rightleftharpoons A^{4+} + 4B^-$
    2. If molar solubility is $S$ (mol/L of $AB_4$ dissolved), then:

    $[A^{4+}] = S, \qquad [B^-] = 4S$

    3. Solubility product expression:

    $K_{sp} = [A^{4+}][B^-]^4 = S\times(4S)^4$

    4. Expand $(4S)^4 = 256S^4$:

    $K_{sp} = S\times256S^4 = 256S^5$

    5. Solve for $S$: …

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