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Order of Reaction

In a 0.250 ...

Question

In a $0.250 M$ solution of $N a_{3} P O_{4}$ , the concentration of the sodium ions would be:

0.250 M

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0.750 M

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0.0833 M

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1.99 M

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Solution

The correct option is B $0.750 M$
$N a_{3} P O_{4}$ dissociates as:

$N a_{3} P O_{4} ⇌ 3 N a^{+} + P O_{4}^{3 -}$

$∴$ conc. of $N a^{+}$ would be $= 3 \times 0.250 = 0.750 M$

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0.250 M

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