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.