Question

Soluble barium compounds are poisonous, but barium sulphate is routinely ingested as a suspended solid in a “barium cocktail” to improve the contrast in X-ray images. Calculate the concentration of dissolved barium per litre of water in equilibrium with solid barium sulphate.

Short answer

The concentration of dissolved barium is $1.0\times {10}^{-5}M$.

Step-by-step solution

Solubility Equilibrium for Barium sulphate.

The solubility equilibrium for the ${\text{BaSO}}_{\text{4}}$compound is as follow:

${\text{BaSO}}_{\text{4}}\text{(s)}\rightleftharpoons {\text{Ba}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

The solubility product expression the ${\text{BaSO}}_{\text{4}}$compound is as follows:

${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ba}}^{\text{2 +}}\right]\left[{\text{SO}}_{\text{4}}^{\text{2 -}}\right]$

$Solubilityproduct\left(K_{sp}\right)ofthe{\text{BaSO}}_{\text{4}}compoundis1.1\times {10}^{-10}.$

Substituting the Concentration.

Let $s$be the molar solubility of $BaS{O}_4$since one mole of$BaS{O}_4$forms 1 mole of $B{a}^{+2}$, the molar solubility of $B{a}^{2+}$is since the mole of$BaS{O}_4$forms 1 mole of $S{O^{2-}}_4$, the molar solubility of $S{O^{-2}}_4$is .

The concentration table of $BaS{O}_4$is as follow:

${\text{BaSO}}_{\text{4}}\text{(s)}\rightleftharpoons {\text{Ba}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Substituting the concentration in the $K_{sp}$expression to get,

$$\begin{gathered} 1.1\times {10}^{-10}=s\times s \\ 1.1\times {10}^{-10}=s^2 \end{gathered}$$

$$\begin{gathered} s=\sqrt{1.1\times {10}^{-10}} \\ =1.0\times {10}^{-5} \end{gathered}$$

So, the molar solubility of $BaS{O}_4$is $1.0\times {10}^{-5}M$.

Concentration of dissolved barium.

Solubility equilibrium for the$BaS{O}_4$compound is given by,

${\text{BaSO}}_{\text{4}}\text{(s)}\rightleftharpoons {\text{Ba}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

One mole of${\text{BaSO}}_{\text{4}}$salt produces 1 mole of $B{a}^{2+}$ions.

Thus,

$\left[{\text{Ba}}^{\text{2 +}}\right]\text{=}\left[{\text{BaSO}}_{\text{4}}\right]$

$=s$

$=1.0\times {10}^{-5}M$

Therefore, the concentration of dissolved barium is $1.0\times {10}^{-5}M$.