what is the concentration of sodium ions in 0.325 m na₂co₃?
what is the concentration of sodium ions in 0.325 m na₂co₃?
Answer
Explanation:
Step1: Write the dissociation equation
$$\ce{Na_{2}CO_{3}(aq) -> 2Na^{+}(aq) + CO^{2-}{3}(aq)}$$ From the equation, 1 mole of $\ce{Na{2}CO_{3}}$ dissociates to give 2 moles of $\ce{Na^{+}}$ ions.
Step2: Calculate the concentration of $\ce{Na^{+}}$ ions
If the concentration of $\ce{Na_{2}CO_{3}}$ is $C = 0.325\space M$, and the mole ratio of $\ce{Na_{2}CO_{3}}$ to $\ce{Na^{+}}$ is $1:2$. Using the formula $C_{\ce{Na^{+}}}=2\times C_{\ce{Na_{2}CO_{3}}}$ Substitute $C_{\ce{Na_{2}CO_{3}}}=0.325\space M$ into the formula: $C_{\ce{Na^{+}}}=2\times0.325\space M$
Answer:
$0.650\space M$