Question

What is the molarity of the HNO₃ solution used in the reaction HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l) if 64.0 mL of the solution is needed to react with 25.0 mL of 0.150 M KOH solution?

A) 0.150 M
B) 0.0375 M
C) 0.0586 M
D) 0.075 M

Answer 1

The molarity of the HNO₃ solution is determined by calculating the moles of KOH reacted and using the stoichiometry of the balanced chemical equation. For a one-to-one reaction, the moles of HNO₃ equate to the moles of KOH. The molarity is then calculated as 0.0586 M.

Step-by-step explanation:

To find the molarity of the HNO₃ solution used in the titration with KOH, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between acids and bases, which is usually one-to-one for strong acids and strong bases:

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

Step 1: Calculate the moles of KOH that reacted. Since the molarity and volume of KOH are given:

# moles KOH = Molarity of KOH x Volume of KOH in liters

# moles KOH = 0.150 M x 0.0250 L = 0.00375 moles

Step 2: According to the balanced equation, the mole ratio of HNO₃ to KOH is 1:1.

# moles HNO₃ = # moles KOH = 0.00375 moles

Step 3: Now we can calculate the molarity of the HNO₃ solution using its definition (moles of solute per liter of solution).

Molarity of HNO₃ = # moles HNO₃ / Volume of HNO₃ in liters

Molarity of HNO₃ = 0.00375 moles / 0.0640 L = 0.0586 M

Therefore, the molarity of the HNO₃ solution is 0.0586 M, which corresponds to answer choice C.