Question

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}

Answer 1

`4.4 * 10^0"mol L"^(-1)`

Explanation:

Sodium carbonate, `"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 `color(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 `color(red)(2)` moles of sodium cations and `1` mole of carbonate anions in aqueous solution.

`"Na"_ color(red)(2)"CO"_ (3(aq)) -> color(red)(2)"Na"_ ((aq))^(+) + "CO"_ (3(aq))^(2-)`

This tells you that a sodium carbonate solution will always have

`["Na"^(+)] = color(red)(2) * ["Na"_ 2"CO"_3]`

In your case, the solution will have

`["Na"^(+)] = color(red)(2) * 2.2 * 10^(0)` `"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.