Question

An aqueous solution contains an unknown concentration of $$B{a}^{+2}$$. When 50 mL of a 1 M solution of $$N{a}_{2}S{O}_4$$​ is added, $$BaS{O}_4$$ ​ just begins to precipitate. The final volume is 500 ml. The solubility product of $$BaS{O}_4$$​ is $$1\times {10}^{-10}$$. Find the concentration of $$B{a}^{+2}$$in the original solution?
$$a)1.1\times {10}^{-9}M$$
$$b)1.0\times {10}^{-10}M$$
$$c)5\times {10}^{-9}M$$
$$d)2\times {10}^{-9}M$$

Answer 1

Hint: Apply the equation of dilution law to the given problem to help determine the concentration of the $S{O_4}^{2-}$ ion in the new solution. Then equate the ionisation product of the given chemical substance to the solubility product to help find the answer.

Complete step by step answer:
Let us first understand the concept of solubility product and how it's used for ionic compounds before trying to use these concepts to help solve the given question.
The solubility product constant, $$K_{sp}$$ ​, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the $$K_{sp}$$ value it has.
Consider the general dissolution reaction below (in aqueous solutions):
$$aA\left(s\right)\rightleftharpoons cC\left(aq\right)+dD\left(aq\right)$$
To solve for the $$K_{sp}$$ it is necessary to take the molarities or concentrations of the products (cC and dD) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power (and also multiply the concentration by that coefficient). This is shown below:
$$K_{sp}$$ = $${\left[C\right]}^c{\left[D\right]}^d$$
Let us now apply the dilution law to help find the concentration of the $S{O_4}^{2-}$ ion in the new solution.
$\begin{array}{rl}& M_1{V}_1=M_2{V}_2,\\ & \text{where }{\text{M}}_1\text{ and }{\text{M}}_2\text{ are the molarities of the substance }\\ & \text{in the old and the new solutions respectively }\\ & \text{with }{\text{V}}_1\text{ and }{\text{V}}_2\text{ being their respective volumes}\text{.}\end{array}$
That implies,
$1\times 50=M_2\times 500$
Therefore, we can say that the concentration of the $S{O_4}^{2-}$ ion in the new solution is 0.1M.
Applying this to the equation of solubility product.
$\begin{array}{rl}& [B{a}^{2+}][S{O_4}^{2-}]=K_{sp}\\ & \Rightarrow [B{a}^{2+}]\times 0.1={10}^{-10}\\ & \Rightarrow [B{a}^{2+}]={10}^{-9}\end{array}$
Let us now use this to try and find the old concentration of $$B{a}^{+2}$$ ions by again putting these values into dilution law.
$\begin{array}{rl}& M_1\times 450={10}^{-9}\times 500\\ & \Rightarrow M_1=1.11\times {10}^{-9}\end{array}$

Therefore, we can safely conclude that the answer to this question is a).

Note:
To avoid confusing clutter, solubility product expressions are often written without the state symbols. Even if you don't write them, you must be aware that the symbols for the ions that you write are for those in solution in water.