$20 mL$ of a $0.2 M A l_{2} (S O_{4})_{3}$ solution is mixed with $20 mL$ of a $0.6 M B a C l_{2}$ solution. Determine the molarity of $A l^{3 +}$ ions in the solution.(in mol/L)

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Solution

We know,
$Molarity = \frac{moles of the solute}{total volume of the solution in mL} \times 1000$
Reaction :
$t = 0 3 B a C l_{2} 20 \times 0.6 m i l l i m o l e + A l_{2} (S O_{4})_{3} 20 \times 0.2 m i l l i m o l e \to 3 B a S O_{4} - + 2 A l C l_{3} -$
$t = \infty 3 B a C l_{2} 0 + A l_{2} (S O_{4})_{3} 0 \to 3 B a S O_{4} 12 m i l l i m o l e + 2 A l C l_{3} 8 m i l l i m o l e$
$A l^{3 +}$ ions present in solution which is only belongs to $A l C l_{3}$ and volume of solution is 40 mL.

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

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