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

When equal volumes are mixed, the molarity of each component will be halved. 
Let’s calculate the value of \( Q \) for precipitation and compare it with \( K_{sp} \). For precipitation, we use the equation: \[ Q = [A^{1-}][Y^{1-}] \] Let’s calculate \( Q \) for each option: \[ Q = (1.8 \times 10^{-7}) \left( \frac{5 \times 10^{-4}}{2} \right)^2 \] This will give: \[ Q = (1.8 \times 10^{-7}) \times (2.5 \times 10^{-4})^2 = 1.8 \times 10^{-7} \times 6.25 \times 10^{-8} = 1.125 \times 10^{-14} \] We can see that the value of \( Q \) for this combination is smaller than \( K_{sp} \), indicating that a precipitate will form. Now let’s check the other combinations for comparison: \[ Q = (10^{-7}) \left( \frac{0.4 \times 10^{-3}}{2} \right)^2 \] \[ Q = (10^{-7}) \times (0.2 \times 10^{-3})^2 = 10^{-7} \times 4 \times 10^{-8} = 4 \times 10^{-15} \] Again, this \( Q \) value is smaller than \( K_{sp} \), indicating no precipitate.