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

90 mL

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.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8