Which one are correct about the solution that contains $3.42 p p m$ $A l_{2} (S O_{4})_{3}$ and $1.42 p p m$ $N a_{2} S O_{4}$ ?

[A l^{3 +}] = [N a^{+}]

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[S O_{4}^{2 -}] = [N a^{+}] = [A l^{3 +}]

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[S O_{4}^{2 -}] = [N a^{+}] + [A l^{3 +}]

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[S O_{4}^{2 -}] = [N a^{+}]

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Solution

The correct options are
A $[A l^{3 +}] = [N a^{+}]$
C $[S O_{4}^{2 -}] = [N a^{+}] + [A l^{3 +}]$
ppm = number of milligrams present in 1 litre of the solution

Given
Mass of $A l_{2} (S O_{4})_{3} = 3.42 m g$
Mass of $N a_{2} S O_{4} = 1.42 m g$

$[A l_{2} (S O_{4})_{3}] = \frac{3.42 \times 10^{- 3}}{342} = 0.01 \times 10^{- 3} M$
$[S O_{4}^{2 -}] = 0.03 \times 10^{- 3} M$
$[A l^{3 +}] = 0.02 \times 10^{- 3} M$

$[N a_{2} S O_{4}]$ $= \frac{1.42 \times 10^{- 3}}{142} = 0.01 \times 10^{- 3} M = [S O_{4}^{2 -}]$
$[N a^{+}] = 0.02 \times 10^{- 3} M$
total concentration of $[S O_{4}^{2 -}]$ $= 0.04 \times 10^{- 3} M$
Hence in solution
$[N a^{+}] = 0.02 \times 10^{- 3} M$
$[S O_{4}^{2 -}]$ $= 0.04 \times 10^{- 3} M$
$[A l^{3 +}] = 0.02 \times 10^{- 3} M$
So,
$[A l^{3 +}] = [N a^{+}]$ and $[S O_{4}^{2 -}] = [N a^{+}] + [A l^{3 +}]$

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Similar questions

$N a^{+} . M g^{2 +}, A l^{3 +}$ and $S i^{4 +}$ are isoelectronic. The order of their ionic size is

N a^{+}

M g^{2 +}

A l^{3 +}

S i^{4 +}

A l^{3 +}, N a^{+}, C a^{2 +}

N a^{+} . A l^{3 +}

B a^{2 +}

N a^{+}, A l^{3 +}

B a^{2 +}

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