HCl(aq) + NaOH(aq) ––––> NaCl(aq) + H2O(l) What volume of 0.631 M HCl is required to react with 15.8 mL of 0.321 M NaOH?

The correct answer is 8.04 mL

Explanation:

Given the neutralization reaction:

HCl(aq) + NaOH(aq) ––––> NaCl(aq) + H₂O(l)

1 mol of HCl reacts with 1 mol of NaOH.

The number of moles of NaOH there is in 15.8 mL is calculated as follows:

0.321 mol/L x 1 L/1000 mL x 15.8 mL = 5.07 x 10⁻³ mol NaOH

Thus, we need 5.07 x 10⁻³ mol of HCl to react with 5.07 x 10⁻³ mol NaOH. We have a 0.631 M HCl solution. We calculate the volume of HCl we need by considering that 1 mol NaOH reacts with 1 mol HCl and that there are 0.631 moles of HCl in 1 liter of solution (1000 mL):

5.07 x 10⁻³ mol NaOH x 1 mol HCl/1 mol NaOH x 1000 mL/0.631 mol HCl = 8.04 mL