If equal volumes of AB and XY (both are salts) aqueous solutions… — Solution | JEE Chemistry

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If equal volumes of AB and XY (both are salts) aqueous solutions are mixed, which of the following combination will give precipitate of AY, at $300 \, \text{K}$ ?

A
$K$ ( $300 \, \text{K}$ ) for AB = $5.2 \times 10^3$
B
$K$ ( $300 \, \text{K}$ ) for AB = $1.0 \times 10^3$
C
$K$ for $10^{-10} \, \text{M} \, AB$ , $5 \times 10^{-10} \, \text{M} \, XY$
D
$K$ for $15 \times 10^{-10} \, \text{M} \, XY$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Equal volumes of aqueous solutions of AB and XY are mixed. We need to identify the combination that gives a precipitate of AY at $300 \, \text{K}$ .

Find: Which option satisfies the precipitation condition for AY.

For a precipitate of AY to form, the ionic product must exceed the solubility product.

\text{AB(aq)} \rightarrow \text{A}^+\text{(aq)} + \text{B}^-\text{(aq)}

\text{XY(aq)} \rightarrow \text{X}^+\text{(aq)} + \text{Y}^-\text{(aq)}

\text{A}^+\text{(aq)} + \text{Y}^-\text{(aq)} \rightleftharpoons \text{AY(s)}

The ionic product is

Q = [\text{A}^+][\text{Y}^-]

Precipitation occurs when

Q > K_{sp}(\text{AY})

Using Equal-Volume Mixing

When equal volumes of the two solutions are mixed, the total volume becomes double, so each initial concentration is halved.

If the initial concentrations are $C_{AB}$ and $C_{XY}$ , then after mixing,

[\text{A}^+] = \frac{C_{AB}}{2}, \qquad [\text{Y}^-] = \frac{C_{XY}}{2}

Hence,

Q = \left(\frac{C_{AB}}{2}\right)\left(\frac{C_{XY}}{2}\right) = \frac{C_{AB} \times C_{XY}}{4}

Check the Complete Numerical Option

Among the listed options, the solution identifies option C as the correct choice and treats it as the complete concentration-based combination.

For option C:

C_{AB} = 1 \times 10^{-10} \, \text{M}, \qquad C_{XY} = 5 \times 10^{-10} \, \text{M}

So,

Q = \frac{(1 \times 10^{-10})(5 \times 10^{-10})}{4}

Q = \frac{5 \times 10^{-20}}{4} = 1.25 \times 10^{-20} \, \text{M}^2

Therefore, precipitation of AY will occur if

K_{sp}(\text{AY}) < 1.25 \times 10^{-20} \, \text{M}^2

The extracted solution explicitly concludes that the correct option is C. Therefore, the correct option is C.

Common mistakes

Using the initial concentrations directly in $Q = [\text{A}^+][\text{Y}^-]$ after mixing is incorrect because equal volumes are mixed, so each concentration becomes half. Use $\frac{C_{AB}}{2}$ and $\frac{C_{XY}}{2}$ before forming the ionic product.
Comparing the ionic product with the wrong criterion is a common error. A precipitate forms only when $Q > K_{sp}$ , not when $Q < K_{sp}$ . Always check the inequality in the correct direction.
Ignoring that the options are fragmented can lead to overinterpreting incomplete data. The provided solution treats option C as the meaningful concentration-based combination, so the conclusion should be grounded in that extracted working rather than invented missing values.

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Answer

Step-by-step solution

Common mistakes

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