C Exercise A student titrated a solution c point using 2894

C. Exercise A student titrated a solution c point using 28.94 ml of 0.3021 M KOH solution. What is the molar mass of the unknow acid? Hint: you must write a balanced equation for the reaction Setup: ontaining 3.7066 g of an unknown triprotic acid to the end

Solution

   H3A(aq) + 3KOH(aq) ---> K3A(aq) + 3H2O(l)

1 mol H3A = 3 mol KOH

No of mol of KOH = 28.94*0.3021 = 8.74 mmol

No of mol of triprotic acid = x mmol

No of mol of triprotic acid = 8.74/3 = 2.913 mmol

Molarmass of triprotic acid = w/n

                           = 3.7066/(2.913*10^-3)

                           = 1272.434 g/mol