A 0.120 g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1218 M NaOH. The equivalence point is reached after adding 12.4 mL of base.
What is the molar mass of the unknown acid?
To determine the molar mass of the unknown acid, we can use the concept of equivalence point and stoichiometry.
First, let's calculate the number of moles of NaOH that reacted with the diprotic acid. The balanced chemical equation for the reaction of NaOH and diprotic acid is as follows:
H2A + 2NaOH -> 2H2O + Na2A
From the equation, we can see that 2 moles of NaOH react with 1 mole of diprotic acid (H2A).
Given that the volume of NaOH used at the equivalence point is 12.4 mL or 0.0124 L, we can calculate the number of moles of NaOH:
moles of NaOH = concentration of NaOH * volume of NaOH
= 0.1218 M * 0.0124 L
= 0.00150832 moles NaOH
Since the mole ratio of NaOH to H2A is 2:1, the number of moles of H2A (diprotic acid) is half the number of moles of NaOH:
moles of H2A = 0.00150832 moles NaOH / 2
= 0.00075416 moles H2A
Next, we can determine the molar mass of the diprotic acid using the mass and moles:
molar mass (M) = mass / moles of H2A
= 0.120 g / 0.00075416 moles H2A
= 159.08 g/mol
Therefore, the molar mass of the unknown diprotic acid is approximately 159.08 g/mol.
H2A + 2NaOH ==> Na2A + 2H2O
mols NaOH = M x L = ?
Using the coefficients in the balanced equation, convert mols NaOH to mols H2A (that is 1/2 mols NaOH)
Then mols = grams/molar mass. You know grams and mols, solve for molar mass.