Question

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 K?

• A. \( K \) (300 K) for \( AB \) = \( 5.2 \times 10^3 \)
• B. \( K \) (300 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 \( 15 \times 10^{-10} \, \text{M} \, XY \)

Hint:

When dealing with solubility products, remember that the concentration of ions is key. Use the product of ion concentrations and compare it with \( K_{sp} \) to determine if a precipitate will form.

Answer 1

The Correct Option is C

The problem asks to identify which combination of aqueous salt solutions AB and XY, when mixed in equal volumes, will result in the formation of a precipitate of the salt AY. The provided options are fragmented, so we will proceed by analyzing the chemical principles and applying them to the most complete and plausible option.

Concept Used:

For a precipitate of a sparingly soluble salt like AY to form, the Ionic Product (Q) of its constituent ions in the solution must exceed its Solubility Product (K_sp).

The dissociation reactions are:

\[ \text{AB(aq)} \rightarrow \text{A}^+\text{(aq)} + \text{B}^-\text{(aq)} \] \[ \text{XY(aq)} \rightarrow \text{X}^+\text{(aq)} + \text{Y}^-\text{(aq)} \]

The precipitation reaction is:

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

The Ionic Product for AY is given by \( Q = [\text{A}^+][\text{Y}^-] \).

The condition for precipitation is:

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

When equal volumes of two solutions are mixed, the final volume is doubled, and the concentration of each solute is halved.

Step-by-Step Solution:

Step 1: Formulate the expression for the Ionic Product after mixing.

Let the initial concentration of the AB solution be \( C_{AB} \) and that of the XY solution be \( C_{XY} \). When equal volumes are mixed, the new concentrations in the mixture become:

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

The Ionic Product \( Q \) is then:

\[ Q = [\text{A}^+][\text{Y}^-] = \left( \frac{C_{AB}}{2} \right) \left( \frac{C_{XY}}{2} \right) = \frac{C_{AB} \times C_{XY}}{4} \]

Step 2: Analyze the provided options.

The options are poorly formatted. However, one option provides a complete set of initial concentrations: "\(\text{K for } 10^{-10} \, \text{M} \, \text{AB}, 5 \times 10^{-10} \, \text{M} \, \text{XY}\)". We interpret this as a scenario where the initial concentrations are \( C_{AB} = 10^{-10} \, \text{M} \) and \( C_{XY} = 5 \times 10^{-10} \, \text{M} \). For precipitation to occur in this case, the \( K_{sp} \) of AY must be smaller than the calculated \( Q \).

Step 3: Calculate the Ionic Product (Q) for the plausible scenario.

Using the initial concentrations from the interpreted option:

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

Now, calculate \( Q \):

\[ Q = \frac{(1 \times 10^{-10}) \times (5 \times 10^{-10})}{4} = \frac{5 \times 10^{-20}}{4} = 1.25 \times 10^{-20} \, \text{M}^2 \]

Step 4: Determine the condition for precipitation.

Precipitation of AY will occur if \( K_{sp}(\text{AY}) \) is less than the calculated Ionic Product \( Q \).

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

Therefore, the combination that will likely give a precipitate is the one specified: \( K \) for \( 10^{-10} \, \text{M} \, AB \), \( 5 \times 10^{-10} \, \text{M} \, XY \), assuming an appropriate \( K_{sp} \) for AY.