Question

A student places 50.0 mL of 0.630 M HNO₂ in a 250mL Erlenmeyer flask and adds indicator. After 35.0 mL of 1.014 M standardized NaOH is added, what is the pH of the solution.

Answer 1

given that a - Solution: Volume of HHOZ (V.)- 50m) Mi = [hing? 0.630M M2= [Maon] 1.014M Volume of Naoh Solution (12) - 35.0ml mmoles of HNO2= molarity X Volume M, XV, = 0.630 x 50 = 31.5 = 31.5 m moles of Naod - molarity x x Volume = M2 X V₂ 1024X35 38.49 . 2

in the reaction Naon + HNO2 Nang ttho at tza mmoles 35.49 31.5 at t time 35.49-31.50 31.5-31.50 - 3.99 -O mmoles 3.99 3.93 molarity of naon a Volume 50+35 85 =0.0469M resultant solution has strong Base Nach So, ph= pon. - loylon ] = -log(0.0469) 1.33 But PH= 14-pon so pH 14- 1.33 = 12.67 pH = 12-67