Question

To neutralize 20.5 ml of a 0.332 m nitric acid solution, you would need 11.9 ml of a 0.178 m potassium hydroxide solution. Is this statement true or false?

Answer 1

The statement is false: 38.24 mL of the 0.178 M KOH solution is required to neutralize 20.5 mL of a 0.332 M HNO3 solution, not 11.9 mL.

Step-by-step explanation:

To determine whether the statement is true, we need to ensure that the mole of HNO3 is neutralized by an equal mole of KOH in the reaction. Nitric acid (HNO3) reacts with potassium hydroxide (KOH) in a 1:1 ratio, according to the equation:

HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l)

First, we calculate the number of moles of HNO3:

Moles of HNO3 = Molarity of HNO3 × Volume of HNO3 (in liters)

Moles of HNO3 = 0.332 mol/L × 20.5 mL × (1 L / 1000 mL) = 0.006806 moles

Next, we calculate the number of moles of KOH needed to neutralize the HNO3:

Moles of KOH = Moles of HNO3 (because of the 1:1 molar ratio)

Moles of KOH = 0.006806 moles

Then, we find the required volume of KOH:

Volume of KOH = Moles of KOH / Molarity of KOH

Volume of KOH = 0.006806 moles / 0.178 mol/L = 38.24 mL

According to this calculation, 38.24 mL of the 0.178 M KOH solution would be needed to neutralize 20.5 mL of a 0.332 M HNO3 solution, not 11.9 mL as stated. Therefore, the statement is false.