Question

What is the molarity of a solution that contains 0.978 grams of H3PO4 in 60.0 mL of solution? What is the new molarity of the H3PO4 solution if the original solution is diluted by a factor of 3? How many mL of the diluted solution could be completely neutralized by 12.5 mL of 0.545 M NaOH?

Answer 1

Molarity of solution(M1) = (w/M)*(1000/V in ml)

                        = (0.978/98)*(1000/60)

           = 0.166 M

volume of solution (V1) = 60 ml

volume of diluted solution(V2) = 60*3 = 180 ml

concentration of diluted solution(M2) =(0.978/98)*(1000/180)

               = 0.0554 M

no of mol of H3PO4 present in solution = 0.978/98 = 0.01 mol

no of mol of NaoH added = 12.5*0.545 = 6.8125 mmol

1 mol H3PO4 = 3 mol NaOH

no of mol of h3po4 reacted = (6.8125/3*10^-3) = 0.00227 mol

volume of dilute H3PO4 reacted = 0.00227*10^3/0.0554

                   = 41 ml