Question

A buffer solution contains 0.34 mol of hypoiodous acid (HIO) and 0.37 mol of sodium hypoiodite (Nalo) in 4.20 L. The K, of hypoiodous acid (HIO) is Ka = 2.3e-11. (a) What is the pH of this buffer? pH = (b) What is the pH of the buffer after the addition of 0.06 mol of NaOH? (assume no volume change) pH = (c) What is the pH of the original buffer after the addition of 0.07 mol of HI? (assume no volume change) pH =

Answer 1

0.34 mol a) Concentrati on of HIO ,HIO 0.081M 4.20 L 0.37 mol Concentrati on of NaIO IO 0.088 M 4.20 L pk of HIOog K =-log (2.3x10-) = 10.64 According to Henderson equation OIT pH pK [но] 0.088 M = 10.64+log = 10.68 0.081 M 0.06 mol b) Concentration of OH =0.014 M 4.20 L ICE table OH OI HO HIO (м): C(M) 0.014 E(M) 0.081 0.014 0.088 -0.014 +0.014 0.067 0 0.102 According to Hen derson equation OI pH pKo [но] 0.102 M = 10.64 log 10.82 0.067 M 0.07 mol c) Concentrati on of H = 0.017 M 4.20 L ICE table Но + Н,О H* OIT I(M) C(M) 0.017 E (M) 0.088 0.017 0.081 -0.017 0.014 0.071 0 0.095 According to Henderson equation РH 3 pК, +10g [но 0.071M = 10.64+log = 10.51 0.095 M