Answer to Question #111144 in General Chemistry for mya 12

Question

At 25 °C, only 0.0140 mol of the generic salt AB3 is soluble in 1.00 L of water.

What is the

Answer 1

Correct task : At 25°C, only 0.0140 mol of the generic salt AB₃ is soluble in 1.00 L of water. What is the K_sp of this salt at 25°C.

Solution:

When the salt dissolves, it dissociates as follows:

AB₃ --> A³⁺ + 3B⁻

--S--------S-------3S--

The molar solubility (S) is the number of moles that can dissolve in 1 L of solution.

Molar solubility of salt is (0.0140 mol) / (1.00 L) = 0.0140 mol/L.

According to the dissociation equation:

Solubility of A³⁺ is 0.0140 mol/L and solubility of B⁻ is 3×0.0140 mol= 0.0420 mol/L.

[A³⁺] = 0.0140 mol/L.

[B⁻] = 0.0420 mol/L.

K_sp is the solubility product constant and calculated as follows:

K_sp(AB₃) = [A³⁺] × [B⁻]³

K_sp(AB₃) = [0.0140] × [0.0420]³

K_sp(AB₃) = 10.37×10^-7.

Ksp of this salt is 1.04×10^-6.

Answer: 1.04×10^-6 is the K_sp of this salt at 25°C.