Question

60.0 mL of 0.0100 M H3PO4 are required to neutralize 30.0 mL of a solution of KOH. What is the molarity of the potassium hydroxide?

Answer 1

To determine the molarity of the potassium hydroxide (KOH) solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between H3PO4 and KOH:

H3PO4 + 3 KOH → K3PO4 + 3 H2O

From the balanced equation, we can see that the stoichiometric ratio between H3PO4 and KOH is 1:3. This means that one mole of H3PO4 reacts with three moles of KOH.

Given information:
Volume of H3PO4 solution = 60.0 mL
Molarity of H3PO4 solution = 0.0100 M
Volume of KOH solution = 30.0 mL

To calculate the moles of H3PO4, we can use the formula:

Moles of solute = Molarity × Volume (in liters)

Moles of H3PO4 = 0.0100 M × (60.0 mL / 1000 mL/L) = 0.0006 moles

Since the stoichiometric ratio between H3PO4 and KOH is 1:3, the moles of KOH will be three times the moles of H3PO4:

Moles of KOH = 3 × 0.0006 moles = 0.0018 moles

To find the molarity of KOH, we can use the formula:

Molarity = Moles of solute / Volume (in liters)

Molarity of KOH = 0.0018 moles / (30.0 mL / 1000 mL/L) = 0.06 M

Therefore, the molarity of the potassium hydroxide (KOH) solution is 0.06 M.

Answer 2

To find the molarity of the potassium hydroxide (KOH) solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between H3PO4 and KOH.

The balanced equation for the reaction is: H3PO4 + 3KOH → K3PO4 + 3H2O

From the balanced equation, we can see that the stoichiometric ratio between H3PO4 and KOH is 1:3. This means that for every 1 mole of H3PO4, we need 3 moles of KOH to react completely.

Given: Volume of H3PO4 solution (V1) = 60.0 mL = 0.0600 L Molarity of H3PO4 solution (M1) = 0.0100 M Volume of KOH solution (V2) = 30.0 mL = 0.0300 L

To find the molarity of KOH, we can use the formula: M1V1 = M2V2

Substituting the given values into the equation, we have: (0.0100 M) × (0.0600 L) = M2 × (0.0300 L)

Solving for M2, the molarity of KOH: M2 = (0.0100 M) × (0.0600 L) / (0.0300 L) M2 = 0.0200 M

Therefore, the molarity of the potassium hydroxide (KOH) solution is 0.0200 M.