A 10.8ml sample of sulfuric acid titrated with 80.0 ml of 0.200 m mg solution. What is the concentration of the sample given the acid-base reaction shown below

1.48 M

Explanation:

Step 1:

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

Mg + H2SO4 —> MgSO4 + H2

Step 2:

Determination of the number of mole of Mg in 80.0 mL of 0.200 M Mg solution. This is illustrated below:

Molarity of Mg = 0.200 M

Volume of solution = 80 mL = 80/1000 = 0.08L

Mole of Mg =?

Molarity = mole /Volume

0.2 = mole /0.08

Mole = 0.2 x 0.08

Mole of Mg = 0.016 mole.

Step 3:

Determination of the number of mole of H2SO4 that reacted. This is illustrated below:

Mg + H2SO4 —> MgSO4 + H2

From the balanced equation above,

1 mole of Mg reacted with 1 mole of H2SO4.

Therefore, 0.016 mole of Mg will also react with 0.016 mole of H2SO4.

Step 4:

Determination of the concentration of the acid.

Mole of H2SO4 = 0.016 mole.

Volume of acid solution = 10.8 mL = 10.8/1000 = 0.0108 L

Molarity =?

Molarity = mole /Volume

Molarity = 0.016/0.0108

Molarity of the acid = 1.48 M

Therefore, the concentration of acid is 1.48 M