# What is the molar concentration of sulfate ions in a 0.150 M Na_2SO_4 solution?

May 28, 2016

#### Answer:

${M}_{S {O}_{4}^{2 -}} = 0.150 M$

#### Explanation:

In solution, sodium sulfate dissociates according to the following equation:

$N {a}_{2} S {O}_{4} \left(s\right) \to 2 N {a}^{+} \left(a q\right) + S {O}_{4}^{2 -} \left(a q\right)$

Therefore, $1 m o l$ of $N {a}_{2} S {O}_{4}$ will give $1 m o l$ of $S {O}_{4}^{2 -}$ and thus, the molarity of sulfate ions will be equal to that of the sodium sulfate.

${M}_{S {O}_{4}^{2 -}} = 0.150 M$

However, the molarity of sodium ions is the double of that of sodium sulfate since $1 m o l$ of $N {a}_{2} S {O}_{4}$ will give $2 m o l$ of $N {a}^{+}$:

${M}_{N {a}^{+}} = 0.300 M$