Explanation:
NaOH (aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Stoichiometric ratio NaOH : HCl = 1:1
moles of HCl needed to be neutralized = 0.18/1000 × 32.5 = 5.85 × 10^-3 mol
So 5.85 × 10^-3 mol of NaOH should be used
If that amount is found in V volume
5.85 × 10^-3 = 0.3/1000 ×V
5.85/0.3 cm^3 = 19.5cm^3
