Question

It takes 120 mL of 0.15 M of carbonic acid (H2CO3) to neutralize 300 mL of sodium hydroxide (NaOH) for the following balanced chemical reaction:
2NaOH + H2CO3 → N2CO3 + 2NaOH

The concentration of the sodium hydroxide _____.

Answer 1

First of all, you need to convert mL to L; 120mL = 0.12L, 300mL = 0.3L.

n (moles of carbonic acid) = C (concentration) x V (volume)
n = 0.15M (0.15 mol L-1) x 0.12L
n = 0.018 mol

C (conc. of NaOH) = 0.018 mol x 2 moles (because there’s 2x NaOH) = 0.036 mol
C = n / V
C = 0.036 mol / 0.3 L
C = 0.12 mol L-1