Chemistry - Concentration Terms Question with Solution | TestHub

Question Explanation: Calculate the concentration of the barium ions in solution.

Concept: Mixing of solution

Solution:

$$\begin{array}{lccc}{\mathrm{BaCl}}_2& +& {\mathrm{K}}_2{\mathrm{CrO}}_4& \to {\mathrm{BaCrO}}_4\downarrow +2\mathrm{KCl}\\ 1\mathrm{L},0.5\mathrm{M}& & 1\mathrm{L},0.2\mathrm{M}& \\ =0.5\mathrm{mol}& & =0.2\mathrm{mol}& (I.R)\end{array}$$

Initially, we have $0.5\mathrm{mol}$ of ${\mathrm{BaCl}}_2$ and $0.2\mathrm{mol}$ of ${\mathrm{K}}_2{\mathrm{CrO}}_4$.

Since the reaction is $1:1$, ${\mathrm{K}}_2{\mathrm{CrO}}_4$ is the limiting reactant.

After the reaction:

Moles of ${\mathrm{BaCl}}_2$ remaining $=0.5\mathrm{mol}-0.2\mathrm{mol}=0.3\mathrm{mol}$.

Moles of ${\mathrm{K}}_2{\mathrm{CrO}}_4$ remaining $=0\mathrm{mol}$.

Moles of ${\mathrm{BaCrO}}_4$ formed $=0.2\mathrm{mol}$.

Moles of $\mathrm{KCl}$ formed $=0.4\mathrm{mol}$.

The total volume of the solution is $1\mathrm{L}+1\mathrm{L}=2\mathrm{L}$.

The concentration of barium ions, ${\mathrm{Ba}}^{2+}$, comes from the remaining ${\mathrm{BaCl}}_2$.

Since ${\mathrm{BaCl}}_2\to {\mathrm{Ba}}^{2+}+2{\mathrm{Cl}}^{-}$, the moles of ${\mathrm{Ba}}^{2+}$ are equal to the moles of ${\mathrm{BaCl}}_2$.

Concentration of ${\mathrm{Ba}}^{2+}=\frac{molesof{\mathrm{Ba}}^{2+}}{totalvolume}=\frac{0.3\mathrm{mol}}{2\mathrm{L}}=0.15\mathrm{M}$.