Question

What volume of 0.300 M NaOH solution is required to neutralize 32.5 cm of a

0.180 M HCl solution?

Answer 1

Step-by-step 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