Question

1)Solid oxalic acid exists as a dihydrate:(COOH)₂*2H₂O.  How many grams of the acid-hydrate was required to make 800.0 ml of a 0.25 M standard oxalic acid solution? (HINT: use 126.07 g/mol as the molecular mass of oxalic acid dihydrate.)

2) Based on the balanced chemical equation presented in your lab manual, what is the molarity of an unknown oxalic acid solution if 10.0 ml of the solution required 28.0 ml of 0.019 M KMnO₄ to reach completion?

Answer 1

1. no of moles of oxalic acid = molarity * volume in L

                                                = 0.25*0.8   = 0.2 moles

mass of oxalic acid                = no of moles * gram molar mass

                                               = 0.2*126.07   = 25.214g

2.    2KMnO4 + 5C2H2O4 +3 H2SO4 ------------------> K2SO4   + 2MnSO4 + 8H2O + 10CO2

       2 moles      5 moles

C2H2O4                                                            KMnO4

M1 =                                                                    M2 = 0.019M

V1 = 10ml                                                            V2 = 28ml

n1 = 5                                                                  n2 = 2

                M1V1/n1      =       M2V2/n2

                     M1          =     M2V2n1/n2V1

                                    =    0.019*28*5/2*10   = 0.133M

molarity of oxalic acid   = 0.133M