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?

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Answer 1

To find the molarity of the potassium hydroxide (KOH) solution, we can use the stoichiometry of the neutralization reaction between H3PO4 and KOH.

The balanced equation for the reaction is:

H3PO4 + 3 KOH → K3PO4 + 3 H2O

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

First, let's calculate the number of moles of H3PO4 used:

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

According to the stoichiometry of the balanced equation, the ratio of H3PO4 to KOH is 1:3. Therefore, the number of moles of KOH used is three times the moles of H3PO4:

Moles of KOH = 3 × 0.0006 moles
Moles of KOH = 0.0018 moles

Next, we calculate the molarity of the KOH solution:

Molarity of KOH = Moles of KOH / Volume (in liters)
Molarity of KOH = 0.0018 moles / (30.0 mL / 1000 mL/L)
Molarity of KOH = 0.06 M

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

Answer 2

Step-by-step explanation:

To determine 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 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 moles of H3PO4, we can use the formula: moles = volume (L) × molarity (M)

Moles of H3PO4 (n1) = V1 × M1 = 0.0600 L × 0.0100 M = 0.00060 mol

According to the stoichiometry of the balanced equation, 1 mole of H3PO4 reacts with 3 moles of KOH. Therefore, the moles of KOH (n2) can be calculated as: n2 = 3 × n1 = 3 × 0.00060 mol = 0.00180 mol

Finally, we can determine the molarity of the KOH solution (M2) using the formula: M2 = moles / volume (L)

Molarity of KOH solution (M2) = n2 / V2 = 0.00180 mol / 0.0300 L = 0.0600 M

Therefore, the molarity of the potassium hydroxide (KOH) solution is 0.0600 M.