How many moles of KOH are needed to exactly neutralize 500. mL of 1.0 M HCl?

How many moles of KOH are needed to exactly neutralize 500. mL of 1.0 M HCl?

• A. 1.0 mol
• B. 2.0 mol
• C. 0.25 mol
• D. 0.50 mol

Answer: (D)

The key idea is that neutralization between a strong acid and a strong base occurs in a 1:1 mole ratio. For HCl and KOH, the reaction is HCl + KOH → KCl + H2O, so the moles of base needed equal the moles of acid present. First, convert the volume of HCl to liters: 0.500 L. Moles of HCl = 1.0 mol/L × 0.500 L = 0.500 mol. Because the reaction is 1:1, you need 0.500 mol of KOH to exactly neutralize the acid. That is 0.50 mol of KOH.

The key idea is that neutralization between a strong acid and a strong base occurs in a 1:1 mole ratio. For HCl and KOH, the reaction is HCl + KOH → KCl + H2O, so the moles of base needed equal the moles of acid present.

First, convert the volume of HCl to liters: 0.500 L. Moles of HCl = 1.0 mol/L × 0.500 L = 0.500 mol. Because the reaction is 1:1, you need 0.500 mol of KOH to exactly neutralize the acid. That is 0.50 mol of KOH.