A solution of a theoretical triprotic acid was prepared by dissolving 5.981 g of solid in...
A solution of a theoretical triprotic acid was prepared by dissolving 5.981 g of solid in enough DI water to make 500.0 mL of solution. 13.87 mL of a 0.446 M solution was required to titrate 20.00 mL of this acid's solution.
What is the concentration of the acid solution?
What is the molar mass of the acid?
Hint: You need to calculate the total moles in the 500.0 mL solution (the full 500.0 mL was NOT titrated).
Homework Answers
Concentration of acid solution = 13.87ml (1L / 500ml) X (0.46 mol / 1L) = 0.0123 mol
Molar mass of the acid = 2.90 g / mol
Moles = mass /MW
MW = mass / moles = 5.981g / 0.0123 mol = 486.2 g/mol
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