An aqueous solution of sodium hydroxide is standardized by titration with a 0.104 M solution of hydrobromic acid. If 13.5 mL of base are required to neutralize 20.2 mL of the acid, what is the molarity of the sodium hydroxide solution?

0.156 M

Explanation:

The neutralization reaction between hydrobromic acid (HBr) and sodium hydroxide (NaOH) is the following:

HBr(ac) + NaOH(aq) → NaBr(aq) + H₂O(l)

As we can see, 1 mol of HBr reacts with 1 mol NaOH. During a titration, at the equivalence point, the number of moles of NaOH added reacts completely with the number of moles of HBr:

moles of NaOH = moles of HBr

The moles of each reactant is calculated as the product of the molarity (M) and the volume (V):

M(NaOH) x V(NaOH) = M(HBr) x V(HBr)

We have the following data:

base: V(NaOH) = 13.5 mL

acid: M(HBr) = 0.104 M;  V(HBr) = 20.2 mL

Thus, we calculate M(NaOH) from the equivalent point equation:

M(NaOH) = (M(HBr) x V(HBr))/(V(NaOH) = (0.104 M x 20.2 mL)/(13.5 mL) = 0.1556 M ≅ 0.156 M

the answer is na (sodium)