Question

Solubility product of $BaSO_4$ is $1.5\times {{10}^{-9}}$ . Calculate the solubility of barium sulphate in pure water and in 0.1M of $BaCl_2$.

Answer 1

Hint: Solubility product constant is a kind of equivalent constant for a solid substance dissolved in an aqueous solution. It is represented as $K_{sp}$. To calculate the solubility we need to write the reaction of the dissociation of the solid and then multiply the individual solubilities of the ions formed raised to the power of their coefficients.

Complete step by step answer:
The solubility product $K_{sp}$ is calculated for a solid substance which is dissolved in water. The solubility product value for a compound is dependent on the solubility of the compound. Higher the solubility, greater is the value of solubility product.
Like for reversible reaction,
\[{{\mathcal{B}}_{b}}{{\mathcal{C}}_{c}}\leftrightharpoons bB+cC\]
\[{{K}_{sp}}={{\left[ B \right]}^{b}}\times {{\left[ C \right]}^{c}}\]
So according to the question $BaSO_4$ dissociates to produce $B{{a}^{+2}}$ and $SO_{4}^{-2}$. The chemical representation is given below:
\[BaS{{O}_{4}}\to B{{a}^{+2}}+SO_{4}^{-2}\]

Suppose the solubility of $B{{a}^{+2}}$ and $SO_{4}^{-2}$ individually in pure water be ‘s’ mol per litre
In pure water,
\[{{K}_{sp}}=\left[ B{{a}^{+2}} \right]\left[ SO_{4}^{-2} \right]\]
\[\begin{align}
  & \Rightarrow {{K}_{sp}}=s\times s \\
 & \Rightarrow 1.5\times {{10}^{-9}}={{s}^{2}} \\
 & \Rightarrow s=\sqrt{1.5\times {{10}^{-9}}} \\
 & \Rightarrow s=3.87\times {{10}^{-5}} \\
\end{align}\]

Hence, the solubility of $BaSO_4$ in pure water is $3.87\times {{10}^{-5}}$.
Now we have to find the solubility of $BaSO_4$ in 0.1 M of $BaCl_2$ solution. As we have already assumed that the solubility of $B{{a}^{+2}}$ is ‘s’. Adding $BaSO_4$ in 0.1 M of $BaCl_2$ solution will increase the concentration of $B{{a}^{+2}}$ by 0.1 M. Thus, the concentration of $B{{a}^{+2}}$ in the solution will be $\left[ B{{a}^{+2}} \right]=0.1+s$. The solubility of $SO_{4}^{-2}$ will remain ‘s’ as no other sulphate ions are added to the solution.

Therefore,
\[\begin{align}
 & \left[ B{{a}^{+2}} \right]=0.1+s \\
 & \left[ SO{{4}^{-2}} \right]=s \\
 & {{K}_{sp}}=\left[ B{{a}^{+2}} \right]\times \left[ SO{{4}^{-2}} \right] \\
 & \Rightarrow {{K}_{sp}}=(0.1+s)(s)=0.1s+{{s}^{2}} \\
\end{align}\]
We can neglect the value of ${{s}^{2}}$, since the value of s >> 1. Thus,
\[K_sp=0.1s\]

But it is given in the question that $K_{sp}=1.5\times {{10}^{-9}}$. Therefore,
\[\begin{align}
 & K_{sp}=0.1s \\
 & \Rightarrow 1.5\times {{10}^{-9}}=0.1s \\
 & \Rightarrow s=\dfrac{1.5\times {{10}^{-9}}}{0.1} \\
 & \Rightarrow s=1.5\times {{10}^{-8}} \\
\end{align}\]
Hence the solubility of $BaSO_4$ in pure water is $3.87\times {{10}^{-5}}$ and the solubility of $BaSO_4$ in 0.1 M of $BaCl_2$ solution is $1.5\times {{10}^{-8}}$.

Note: It should be recalled that the solids do not occur in the equilibrium constant equations. The active mass of solid is unity so don’t consider their concentration. But, when a solid is dissolved in water, and it dissociates, the concentrations are taken into consideration for the ionic product calculation and equilibrium constant calculations. Here for the solubility product, we take that the solubility of the ions as equal.