Question

What is the concentration in M of an unknown tri-protic acid (H₃A) if it takes 42.79 mL of 0.108 M KOH to fully titrate a 10.00 mL sample it? Report your answer to 3 sig figs,

Answer 1

Volume of KOH used for titration = 42.79 mL

Concentration of KOH= 0.108 M

Moles of KOH = 42.79 x 0.108 / 1000 = 0.004621 Moles

Moles of tri-protic acid presents =   0.004621 Moles /3 = 0.001540 Moles

Beucase one triprotic acid required three equivalent of KOH.

Volume of  triprotic acid = 10 ml

Concentration of triprotic acid = 0.001540 x 1000 ml / 10 ml = 0.1540 M

Hence Concentration of triprotic acid is 0.1540 M