Question

$20\text{mL}$ of a $0.2\text{M}A{l}_2(S{O}_4{)}_3$ solution is mixed with $20\text{mL}$ of a $0.6\text{M}BaC{l}_2$ solution. Determine the molarity of $A{l}^{3+}$ ions in the solution.

• A. $0.20\text{M}$
• B. $0.25\text{M}$
• C. $10.3\text{M}$
• D. $0.10\text{M}$

Answer 1

The correct option is A $0.20\text{M}$

$\text{Molarity}=\frac{\text{moles of the solute}}{\text{total volume of the solution in mL}}\times 1000$
Reaction :
$\underset{t=0}{}\underset{20\times 0.6millimole}{3BaC{l}_2}+\underset{20\times 0.2millimole}{A{l}_2(S{O}_4{)}_3}\to \underset{-}{3BaS{O}_4}+\underset{-}{2AlC{l}_3}$
$\underset{t=\infty }{}\underset{0}{3BaC{l}_2}+\underset{0}{A{l}_2(S{O}_4{)}_3}\to \underset{12millimole}{3BaS{O}_4}+\underset{8millimole}{2AlC{l}_3}$
$A{l}^{3+}$ ions present in solution which is only belongs to $AlC{l}_3$ and volume of solution is 40 mL.

$\text{Molarity}=\frac{\text{number of moles of}A{l}^{3+}}{\text{total volume of the solution in mL}}\times 1000$
Molarity:
$=\frac{\left(8/1000\right)}{40}\times 1000=0.20\text{M}$