# At a certain temperature, the solubility of barium chromate (BaCrO_4) is 1.8 x 10^-5 mol/L. What is the Ksp value at this temperature?

Oct 27, 2016

${K}_{s p}$ $=$ $\left[B {a}^{2 +}\right] \left[C r {O}_{4}^{2 -}\right]$ $=$ ${\left(1.8 \times {10}^{-} 5\right)}^{2}$

#### Explanation:

We write the solubility expression in this way:

$B a C r {O}_{4} \left(s\right) r i g h t \le f t h a r p \infty n s B {a}^{2 +} + C r {O}_{4}^{2 -}$

Now ${K}_{s p}$ $=$ $\left[B {a}^{2 +}\right] \left[C r {O}_{4}^{2 -}\right]$ $=$ ${\left(1.8 \times {10}^{-} 5\right)}^{2}$ $=$ $3.24 \times {10}^{-} 10.$

Why do we square this value? We were quoted the solubility of barium chromate. Because the solubility represents the concentrations of chromate ion and barium ion.