Question

In a titration of HCl with NaOH, 100mL of the base was required to neutralize 20mL of 5.0 M HCl. What is the molarity of the NaOH?

Answer 1

Let's write the balanced reaction equation:

NaOH + HCl $\rightarrow$ NaCl + H₂O.

You see that 1 molecule of NaOH reacts with 1 molecule of HCl. Therefore, the number of the moles of HCl and NaOH reacted are equal:

$n(HCl) = n(NaOH)$ .

The number of the moles of HCl reacted is (remember to switch mL to L) :

$n(HCl) = cV = 5.0·20·10^{-3} = 0.1$ mol

Therefore, the number of the moles of NaOH used for the neutralisation of 0.1 mol of HCl is:

$n(NaOH) = 0.1$ mol.

The volume of the base was 100 mL. The concentration of the base was:

$c(NaOH) = n/V = \frac{0.1}{100·10^{-3}} = 1$ M.

Answer: the molarity of NaOH was 1 M.