Question

what is the concentration of sodium ions in 0.325 m na₂co₃?

Explanation:

Step1: Analyze the dissociation of \(Na_2CO_3\)

\(Na_2CO_3\) dissociates as \(Na_2CO_3
ightarrow 2Na^{+}+CO_3^{2 -}\). So, 1 mole of \(Na_2CO_3\) gives 2 moles of \(Na^{+}\) ions.

Step2: Calculate the concentration of \(Na^{+}\) ions

If the concentration of \(Na_2CO_3\) is \(C = 0.325\space M\), and using the mole ratio from the dissociation equation \(n=\frac{n_{Na^{+}}}{n_{Na_2CO_3}}=\frac{2}{1}\). Then the concentration of \(Na^{+}\) ions, \(C_{Na^{+}}=2\times C_{Na_2CO_3}\).
Substitute \(C_{Na_2CO_3}=0.325\space M\) into the formula: \(C_{Na^{+}} = 2\times0.325\space M\)

Answer:

\(0.650\space M\)