If 5.12 liters of a 2.75 M phosphoric acid is neutralized by magnesium hydroxide solution of 4.00 M.
How many liters of base were used in this neutralization reaction?

5.28 L

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 5.12L

Molarity of acid (Ma) = 2.75M

Molarity of base (Mb) = 4M

Volume of base (Vb) =.?

Step 2:

The balanced equation for the reaction

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 2

Mole ratio of the base (nB) = 3

Step 3:

Determination of the volume of the base.

This is illustrated below:

MaVa/MbVb = nA/nB

2.75 x 5.12 / 4 x Vb = 2/3

Cross multiply

4 x 2 x Vb = 2.75 x 5.12 x 3

Divide both side by 4 x 2

Vb = (2.75 x 5.12 x 3)/(4 x 2)

Vb = 5.28 L

Therefore, the volume of the base is 5.28 L

The volume of the base Mg(OH)2 used is 5.28 L

Explanation:

Step 1: Data given

Volume of phosphoric acid = 5.12 L

Concentration of phosphoric acid = 2.75 M

concentration of magnesium hydroxide = 4.00 M

Step 2: The balanced equation

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

Step 3: Calculate the volume of magnesium hydroxide needed

a*Cb*Vb = b*Ca*Va

⇒with a = the coefficient of H3PO4 = 2

⇒with Cb = the concentration of Mg(OH)2 = 4.00 M

⇒with Vb = the volume of Mg(OH)2 = TO BE DETERMINED

⇒with b = the coefficient of Mg(OH)2 = 3

⇒with Ca = the concentration of H3PO4 = 2.75 M

⇒with Va = the volume of H3PO4 = 5.12 L

2 * 4.00 * Vb = 3 * 2.75 * 5.12

Vb = (3*2.75 * 5.12) / (2*4.00)

Vb =  5.28 L

The volume of the base Mg(OH)2 used is 5.28 L