# What is the concentration of sodium ions in a (2.2x10^0) mol/L solution of sodium carbonate? (answer to 2 s.d. in mol/L}

May 29, 2017

$4.4 \cdot {10}^{0} {\text{mol L}}^{- 1}$

#### Explanation:

Sodium carbonate, ${\text{Na"_color(red)(2)"CO}}_{3}$, is an ionic compound that, as its name and chemical formula suggest, contains sodium cations and carbonate anions in a $\textcolor{red}{2} : 1$ ratio.

This salt is soluble in water, which means that every time $1$ mole of sodium carbonate dissolves in water, it produces $\textcolor{red}{2}$ moles of sodium cations and $1$ mole of carbonate anions in aqueous solution.

${\text{Na"_ color(red)(2)"CO"_ (3(aq)) -> color(red)(2)"Na"_ ((aq))^(+) + "CO}}_{3 \left(a q\right)}^{2 -}$

This tells you that a sodium carbonate solution will always have

$\left[{\text{Na"^(+)] = color(red)(2) * ["Na"_ 2"CO}}_{3}\right]$

In your case, the solution will have

$\left[{\text{Na}}^{+}\right] = \textcolor{red}{2} \cdot 2.2 \cdot {10}^{0}$ ${\text{mol L}}^{- 1}$

color(darkgreen)(ul(color(black)(["Na"^(+)] =4.4 * 10^0color(white)(.)"mol L"^(-1))))

The answer is rounded to two sig figs.