Question

10.00 mL Nitric acid was titrated with NaOH of 0.100 molarity until the equivalence was reached. 15.2 mL of NaOH were required. Find:

A) Concentration of nitric acid before titration

B) pH of solution before titration

C) pH of solution after 15.1 mL of sodium hydroxide is added

D) pH of solution after 15.3 mL of sodium hydroxide is added

Answer 1

Let us assume the concentration of HNO2 = (M) so, considering low of equivalence X X lomba 010 x 15.2 mL => x 0.100 x 15.2 la (M) -0.152 (M) 0 concentration of two be forse titration & d. 152 (M) ☺ [H+] = [H NG] (Because Hongz is a strong acid) = 0.152 so, pH before titration = -log [ht] a-log (0.152) -0.818 © After addition of 15.1 ml NaOH of o. 100 (M) Mole of Nach added = D.loo x15131-51x16?male Initial mole of HNO3 = 0.152 ( x 10 = 1.52 x 10 mole So, now, mole of Hng left = (1-52-1.5172103 mole =dial xto? male = [H+] So, pH = -log [**] = o-loy (10-5) = 5 o A For 15.3 mL Nach, more of NaOH = 0.150 xls 3 mole = 1.53810 3 mole so. Mole of excess Nadt= (1.53 -1.52) x103 -0.010103 mole su, poH = -log Colt"]= -log (16-5) = 5 Sy pH = 14-0H=14-5= 9