Question

What is the solubility of AgI in g/L in a 0.040 M solution of MgI2? Ksp for AgI is 8.3 x 10^-17

Answer 1

Answer on the question #49132, Chemistry, Other

Question:

What is the solubility of AgI in g/L in a 0.040 M solution of MgI2? Ksp for AgI is $8.3 \times 10^{-6}$?

Solution:

The equation of AgI dissociation is: $(c(MgI2) = \frac{1}{2} c(I))$

$$AgI = Ag^+ + I$$ $$\begin{array}{l} c_0 \quad - \quad 0 \quad 0.08 \\ \Delta c \quad - \quad x \quad x \\ [c] \quad - \quad x \quad 0.08 + x \\ K_{sp} = [Ag][I] = 8.3 \times 10^{-17} \\ K_{sp} = x(0.08 + x) \end{array}$$

As $x << 0.04$:

$$K_{sp} = 0.08 \times x = 8.3 \times 10^{-17}$$ $$x = 1.05 \times 10^{-14} = c(Ag^+), \frac{mol}{L}$$

The solubility in g/L:

$$s = \frac{m(AgI)}{V(sol)} = \frac{n(AgI)M(AgI)}{V(sol)} = c(Ag^+) \times M(AgI)$$ $$s = 1.05 \times 10^{-15} \times 234,77 = 2.45 \times 10^{-13} \frac{g}{L}$$

Answer: $2.45 \times 10^{-13} \frac{g}{L}$

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