# Question 2e817

Apr 1, 2015

Dissolution value is the molar solubility of the substance which is the molar concentration of a saturated solution. This is calculated from the ${K}_{s p}$ value.

Let's say you have a solid that is dissolving

${A}_{x} {B}_{y} \left(s\right) \rightarrow x A \left(a q\right) + y B \left(a q\right)$

${K}_{s p} = {\left[A\right]}^{x} {\left[B\right]}^{y}$

$A {g}_{2} C r {O}_{4} \left(s\right) \rightarrow 2 A {g}^{+} \left(a q\right) + C r {O}_{4}^{-} \left(a q\right)$ and

${K}_{s p} = {\left[A {g}^{+}\right]}^{2} \left[C r {O}_{4}^{-}\right]$

The dissolution value will be the concentration of the chromate ion (this will also be the concentration of the solution). Each formula unit of the compound dissociates into 2 silver ions and 1 chromate ion so the concentration of silver ions will be double that.

Mathematically, the ${K}_{s p}$ value

${K}_{s p} = {\left[A {g}^{+}\right]}^{2} \left[C r {O}_{4}^{-}\right]$

${K}_{s p} = {\left[2 \left(3.2 \cdot {10}^{- 5}\right)\right]}^{2} \left[3.2 \cdot {10}^{- 5}\right] = 1.3 \cdot {10}^{- 13}$

Apr 6, 2015

K_"sp" = 1.31 ×10^(-13)

#### Explanation:

K_"sp" = 1.31 ×10^(-13)

Every mole of Ag₂CrO₄ that dissolves forms 2 mol of Ag⁺ and 1 mol of CrO₄²⁻.

K_"sp" = ["Ag"^+]^2["CrO"_4^(2-)] = (6.4 × 10^(-5))^2(3.2 × 10^-5) = 1.31 ×10^(-13)#