help please, how to answer this question

Question

The [OH^-] of a solution of 4.85 mol/L sodium carbonate is

Answer 1

NOTE: I'd like to point out that this is an egregiously high concentration of sodium carbonate. Sodium carbonate's solubility in water at 30ºC is about 30 g per 100 g water, which equates to about 2.83 M. This is a maximum concentration at room temperature. So in reality, this problem presents a super saturated solution that is extremely basic. I complete the slightly more "realistic" calculation at the end.

GIVEN:

Before tackling concentration, we need to know what's happening to the sodium carbonate. When we first look at the compound (Na₂CO₃), we're like, where even IS the OH^-?? Let me show you.

When dissolved in water, sodium carbonate creates sodium hydroxide and carbonic acid. It looks like this:

Na₂CO₃ + 2 H₂O → 2 NaOH + H₂CO₃

Now we see the OH taking form on the right side. Since sodium carbonate is highly soluble in water, we will assume (unless you're told otherwise - see below) that 100% of the sodium carbonate has dissociated (i.e. all 4.81 moles in this liter).

However, look at the coefficients in our balanced equation. For every 1 molecule of sodium carbonate, we've created TWO molecules of sodium hydroxide. Therefore, since our eyes are on OH^-, we need to double the number of moles.

4.81 x 2 = 9.62 moles NaOH

So, we actually create 9.62 moles of NaOH from our original 4.81 of sodium carbonate. Since NaOH dissociates completely in water, we have 9.62 moles of Na⁺ and 9.62 moles of OH^-. Therefore, the concentration is:

9.62 moles / 1 L = 9.62 M OH^-.

REALISTIC:

Let's assume they know the concentration is wack and we're smart enough to catch it. We dump a bunch of sodium carbonate in water and end up with a filtered 2.83 M sodium carbonate solution. The same steps would follow, justifying that 1 molecule of sodium carbonate creates 2 NaOH molecules, doubling our concentration. In this case, our OH^- concentration would be 2.83 x 2 = 5.66 M [OH^-].

Regardless, these are extremely high hydroxide concentrations. Please correct me if I've neglected any information.