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.        

        HA  +  NaOH  à  NaA  +  H₂O

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

        Moles of NaOH used in the  titration = C x V,   = 0.0950  x  0.0465    =  0.00442 moles

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

        1:1 ratio, so 0.00442 moles of acid used in the 25cm³ titration

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

        0.00442 moles in 25cm³, so x 10 ( /25, x250) = 0.0442 moles in 250cm³ (of which the 3.88g was dissolved)

e)  Calculate the relative molecular mass of the acid.

        Mr = mass / moles,  3.88 / 0.0442  =  87.8 g mol^-1

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.

        Moles of NaOH used in the  titration = C x V,   = 0.160  x  0.0156    =  0.00250 moles

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

        2:1 ratio of NaOH : acid, so 0.00125 moles of acid used in the 25cm³ titration

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

        0.00125 moles in 25cm³, so x 10 ( /25, x250) = 0.0125 moles in 250cm³ (of which the 1.575g was dissolved)

d)  Calculate the relative molecular mass of ethanedioic acid.

        Mr = mass / moles,  1.575 / 0.0125  =  126 g mol^-1

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

        H₂C₂O₄ has an Mr = 90, 126 - 90 = 36.  36 is the Mr of all the H₂O's.  Each H₂O has an Mr of 18.  36 / 18 = 2.  So n = 2:  H₂C₂O₄.2H₂O

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

        2HCl  +  M₂CO₃  à  2MCl  +  H₂O  +  CO₂  (M must be a 1+ ion as CO₃^2- has a 2- charge.  1+ ions are found in Gp1 so must be a Gp1 metal

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

        Moles of HCl used in the  titration = C x V,   = 0.100  x  0.0270    =  0.00270 moles

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

        2:1 ratio of HCl : M₂CO₃, so 0.00135 moles of M₂CO₃ used in the 25cm³ titration.

        so x 40 ( /25, x1000) M₂CO₃ = 0.0540 moles in 1000cm³ (of which the 7.46g was dissolved)

d)  Calculate the relative molecular mass of M₂CO₃.

        Mr = mass / moles,  7.46 / 0.0540  =  138.1 g mol^-1

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

        CO₃ has an Mr = 60, 138.1 - 60 = 78.1.  78.1 is the Ar of M x 2.  Each M has an Ar of 39.1 (78.1 / 2).  M must be potassium, K