Standard solutions and titration calculations

1)     3.88 g of a monoprotic acid was dissolved in water and the solution made up to 250 cm³.  25.00 cm³ of this solution was titrated with 0.0950 mol dm^-3 NaOH solution, requiring 46.50 cm³. 

a)  Write the chemical equation using HA as the formula of the monoprotic acid.

b)  Calculate the moles of NaOH   used in the titration.

c)  Calculate the number of moles of the monoprotic acid used in the titration.

d)  Use your answer in (c) to calulate the number of moles of the monoprotic acid in the 250cm³ solution.

e)  Calculate the relative molecular mass of the acid.

2)     A 1.575 g sample of ethanedioic acid crystals, H₂C₂O₄.nH₂O, was dissolved in water and made up to 250 cm³.  One mole of the acid reacts with two moles of NaOH.  In a titration, 25.00 cm³ of this solution of acid reacted with exactly 15.60 cm³ of 0.160 mol dm^-3 NaOH.

a)  Calculate the moles of NaOH   used in the titration.

b)  Calculate the number of moles of ethanedioic acid used in the titration.

c)  Use your answer in (b) to calulate the number of moles of ethanedioic acid in the 250cm³ solution.

d)  Calculate the relative molecular mass of ethanedioic acid.

e)  Use your answer to (d) and the formula H₂C₂O₄.nH₂O to calulate the value of n.

3)     A solution of a metal carbonate, M₂CO₃, was prepared by dissolving 7.46 g of the anhydrous solid in water to give 1000 cm³ of solution.  25.00 cm³ of this solution reacted with 27.00 cm³ of 0.100 mol dm^-3 hydrochloric acid. 

a)  Write the chemical equation for the reaction

b)  Calculate the moles of HCl used in the titration.

c)  Calculate the number of moles of M₂CO₃ used in the titration.  Hence calculate the number of moles in 7.46g

d)  Calculate the relative molecular mass of M₂CO₃.

e)  Use your answer to (d) and the Mr of M₂CO₃ and identify the metal M.