50.0 ml of 2.60 M HBr(aq) was exactly neutralized (final pH=7) by a 1.80 M solution of NaOH(aq). How much volume of of the NaOH solution was added ?

72.2 ml

Explanation:

The neutralization equation between HBr (acid) and NaOH (base) is the following:

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

We can see that 1 mol of HBr reacts with 1 mol of NaOH. At the equivalence point, the total number of moles of HBr reacts with the total number of moles of NaOH. The number of moles can be calculated as the product between the molarity (M, in mol/L) and the volume (V). So, we can equal the quantities of acid and base, as follows:

moles HBr = moles NaOH

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

Now, we calculate the volume of NaOH with the data:

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

              = (2.60 mol/L x 50.0 ml)/(1.80 mol/L)

              =72.2 ml

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