Question

How many milliliters of a 0.100 M NaOH solution are needed to neutralize 15.0 mL of 0.200 M H3PO4?

Answer 1

To neutralize 15.0 mL of 0.200 M H3PO4, 90.0 mL of a 0.100 M NaOH solution is required, as determined by stoichiometric calculations based on the balanced chemical equation for the neutralization reaction.

Step-by-step explanation:

To determine how many milliliters of a 0.100 M NaOH solution are needed to neutralize 15.0 mL of 0.200 M H3PO4, we'll need to use a stoichiometric calculation involving the balanced chemical equation for the neutralization reaction.

The balanced equation for the reaction between NaOH and H3PO4 is:

3 NaOH + H3PO4 → Na3PO4 + 3 H2O

The stoichiometry of this reaction indicates that it takes 3 moles of NaOH to react with 1 mole of H3PO4. To find the volume of NaOH needed, we perform the following steps:

Calculate the moles of H3PO4: (0.200 M) × (0.015 L) = 0.003 moles.

Determine moles of NaOH needed using the stoichiometry of the reaction: 3 moles of NaOH per 1 mole of H3PO4 gives 3 × 0.003 = 0.009 moles of NaOH.

Calculate volume of NaOH solution: 0.009 moles ÷ 0.100 M = 0.090 L, which is 90.0 mL.

Therefore, 90.0 mL of a 0.100 M NaOH solution is required to neutralize 15.0 mL of 0.200 M H3PO4.

Answer 2

90 mL

Step-by-step explanation:

Step 1:

Data obtained from the question. This include the following:

Volume of base, NaOH (Vb) =..?

Molarity of base, NaOH (Mb) = 0.1 M

Volume of acid, H3PO4 (Va) = 15 mL

Molarity of acid, H3PO4 (Ma) = 0.2 M

Step 2:

The balanced equation for the reaction. This is given below:

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

From the balanced equation above,

Mole ratio of the acid, H3PO4 (nA) = 1

Mole ratio of base, NaOH (nB) = 3

Step 3:

Determination of the volume of the base. The volume of the base needed for the reaction can be obtained as follow:

MaVa / MbVb = nA/nB

0.2 x 15 / 0.1 x Vb = 1/3

Cross multiply

0.1 x Vb = 0.2 x 15 x 3

Divide both side by 0.1

Vb = (0.2 x 15 x 3) / 0.1

Vb = 90 mL

Therefore, 90 mL of the base, NaOH is needed for the reaction.