0.352 g sample of a diprotic acid is dissolved in water and titrated with 0.150 m naoh.0.150 m naoh. what is the molar mass of the acid if 36.4 ml36.4 ml of the naohnaoh solution is required to neutralize the sample? assume the volume of naohnaoh corresponds to the second equivalence point.

128.94 g/mol

Given;

Mass of diprotic acid, H₂X =0.352 g

Molarity of NaOH = 0.150 M

Volume of the NaOH = 36.4 mL

We are required to calculate the molar mass of the acid.

Note: A diprotic acid is an acid that contains 2 replaceable hydrogen atoms

The balanced equation for the reaction between the diprotic acid and NaOH will be;

H₂X(aq) + 2NaOH(aq) → Na₂X(aq) + 2H₂O(l)

Given the molarity and volume of NaOH we can calculate the number of moles;

Moles of NaOH = Molarity × Volume

                          = 0.150 M × 0.0364 L

                         = 0.00546 moles

From the equation;

1 mole of the acid reacts with 2 moles of NaOH

Therefore; H₂x : NaOH = 1 : 2

Moles of H₂x = 0.00546 moles ÷ 2

                      = 0.00273 moles

Molar mass is the mass equivalent to 1 mole of a compound or element

From our calculations;

0.00273 moles = 0.352 g of the acid;

Therefore, mass in 1 mole ;

= 0.352 g ÷ 0.00273 moles

= 128.94 g/mol

Thus, the molar mass of the diprotic, H₂X is 128.94 g/mol

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun