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Molarity

A piece of Al...

Question

A piece of $A l$ weighing $27 g$ is reacted with $200 mL$ of $H_{2} S O_{4}$ (specific gravity = $1.8$ and $54.5$ by weight). After the metal is completely dissolved, $73 g$ $H C l$ is added and solution is further diluted to $500 mL$ solution. Find the concentration of $H^{+}$ ion in mol/litre.

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Solution

Molarity of $H_{2} S O_{4}$ $= \frac{1.8 \times 54.5 \times 10}{98} = 10 M$

$2 A l 1 mole + 3 H_{2} S O_{4} 2 mole \to A l_{2} (S O_{4})_{3} + 3 H_{2}$
Here $A l$ is limiting reagent.
Moles of $H_{2} S O_{4}$ left $= 2 - 1.5 = 0.5$
Moles of $H C l$ added $= 2$
Final volume of the solution $= 500 m L$
Moles of $H^{+}$ ion $= 3$
Concentration of $H^{+}$ $= 6 M$

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