How many milliliters of a 0.40 M solution of hydrochloric acid are necessary to neutralize 25 mL of a 0.20 M sodium hydroxide solution?

a. 12.5 mL

b. 50 mL

c. 10.5 mL

d. 25 mL

e. 20 mL

option a. 12.5 ml

Explanation:

1) Chemical equation:

This is the neutralization of a strong acid (HCl) and a strong base (NaOH), which is ruled by this equation:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

2) Mole ratio:

1 mol HCl : 1 mol NaOH

3) Number of moles of NaOH

Molarity formula: M = n / V (in liters) Solve for n: n = M*V Substitute values and compute: n = 0.20 M * 0.025 liter = 0.005 moles NaOH.

4) Number of moles of HCl:

From the mole ratio, the number of moles of HCl equals the number of moles of NaOH: 0.005 moles HCl

5) Volume solution HCl:

Molarity formula: M = n / V (in liters) Solve for V: V = n / M Substitute values and compute: V = 0.005 moles / 0.40 M = 0.0125liter Convert to ml: 0.0125 liter * 1000 ml / liter = 12.5 ml

Conclusion: 12.5 ml of a 0.40 M solution of HCl are necessary to neutralize 25 ml of a 0.20 M NaOH solution.