Question

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

Answer 1

A

Step-by-step explanation:

Answer 2

• option a. 12.5 ml

Step-by-step 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.