Question

Q. #11. 25.00 mL of 0.100 M HNO, acid is titrated with 0.110 M sodium hydroxide solution (a) What is the pH of the acid solution before any base solution is added? (b) What is the pH of the titration mixture after 15.00 mL of base are added? (c) What is the pH at the equivalence point/end point?

Answer 1

HNO3 → Shong acid MaoH f Strong base HNOz (aq) + Naom (agyat NaNO3 (94) + Hy grey 25mL of Dolo0 M HNO3 = 25 m** 0-100 bolet' = 215 mmol of HNO @ Initially there was only 25mL of 0.100 M HNO₂ , [HⓇ] = [hnog) = 0.100 M So, ptt = -had, CH®) = -1), (0.100)= 1 6 15 mL of o.llom NaOH = 15 m/x00110 molla = 1.65 mmol of Noou Now, 1 mol of Naot reacts with imot of HNO₃ 1. So, 1165 mol of Nauh will react with 1.65 mmol of thos fo, a movent of encek HNDE = 205 mmol-165 mmol = 0.85 mmol of HANOZ Total volume of the solution = 25mL + 15 mL = 40 mL so, (HNO3) = 0.25mmol 40 L + [he] = ['HNO3) = 0.02.125 M ts, pH = -logy, LH @J = -logo (0-02125) = 1.672641066 © since, both are strong and f strong bases - at equivalence point there will be only Nano, k H₂O. So, at the equivalence point, the solution will be neutral. - 0.02 125 M + pH=7