Question

What volume of 0.325 M KOH can be neutralized with 72.5 of 1.77 M HNO3

Answer 1

To calculate the volume of the KOH solution needed to neutralize the HNO3 solution, we can use the equation: KOH(aq) + HNO3(aq) → KNO3(aq) + H2O(l). Given the molarity of KOH and the volume of HNO3, we can use the molarity-volume relationship to determine the volume of KOH solution. The volume of KOH solution needed is approximately 0.393 L.

Step-by-step explanation:

To calculate the volume of the KOH solution needed to neutralize the HNO3 solution, we can use the equation:

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

Given:

• [KOH] = 0.325 M
• Volume of KOH solution = ??
• [HNO3] = 1.77 M
• Volume of HNO3 solution = 72.5 mL
• = 0.0725 L

We can use the equation:

Molarity x Volume = Moles of solute

For the KOH solution:

(0.325 M) x Volume of KOH solution = Moles of KOH

For the HNO3 solution:

(1.77 M) x 0.0725 L = Moles of HNO3

Since the reaction is 1:1, the moles of KOH will be equal to the moles of HNO3. Therefore:

(0.325 M) x Volume of KOH solution = (1.77 M) x 0.0725 L

Solving for Volume of KOH solution:

Volume of KOH solution = ((1.77 M) x 0.0725 L) / (0.325 M)

Volume of KOH solution = 0.393 L

So, approximately 0.393 L of the 0.325 M KOH solution can be neutralized with 72.5 mL of 1.77 M HNO3 solution.