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Question

$p \times 10^{- q}$ is the molarity of $S O_{4}^{2 -}$ ion in an aqueous solution that contains $34.2 p p m$ of $A l_{2} (S O_{4})_{3}$ . Calculate $(p + q)$ ?
(Assume complete dissociation and density of solution $1 g L^{- 1})$

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Solution

We know, $34.2 p p m$ of $A l_{2} (S O_{4})_{3}$ is actually equal to $34.2 m g$ of $A l_{2} (S O_{4})_{3}$ in $1 L$ solution.
Again, molar mass of $A l_{2} (S O_{4})_{3} = 342 g / m o l$
$∴ Molarity of A l_{2} (S O_{4})_{3} = \frac{\frac{34.2 \times 10^{- 3}}{342}}{1} M = 10^{- 4} M$
$A l_{2} (S O_{4})_{3} (a q) \to 2 A l^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) 10^{- 4} M 2 \times 10^{- 4} M 3 \times 10^{- 4} M [S O_{4}^{2 -}] = 3 \times 10^{- 4} M$
According to the question,
$p = 3 a n d q = 4$
$∴ p + q = 3 + 4 = 7$

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