Question

U t ents ... Evaluation 2 Finding pH Due in: What is the pH of a solution prepared by adding 2.319 of NaNO2 to 253.00 ml of water? (Base Ionization Constant for NaNO, - 2.22E-11) 8.23 0] 1.16 27 Submit Answer Tries 0/98 This discussion is closed. EL Type here to search SUS ل الالالالا/// اااالالالالالالا الاااعاالالا // // /

Answer 1

Amount of NaNO2 added = 2.31 g

Amount of water = 253.00 mL = 0.253 L

Molarity of NaNO2 solution = (2.31/69)/0.253 = 13.23 M

Kb = 2.2 x 10-11

NO2- + H2O ----> HNO2 + OH−

Iintial 13.23 0 0 0

Equil. 13.23-x 0 x x

Kb = (x) (x)/ (13.23-x)

2.2 x 10-11 = x2/13.23

x2 = 2.91 x 10-10

x = 1.705 x 10-5 = [OH-]

[H+] = Kw/[OH-] = 1.0 x 10-14/1.705 x 10-5 = 0.586 x 10-9 = 5.86 x 10-8

pH = -log [H+] = - log 5.86 x 10-8 = 8- log 5.86 = 8 - 0.767 = 7.23 ≈ 7.27