Question

What is the pH of 0.40 M Na2SO3? (K1 for H2SO3 = 1.5 x 10^-2; K2 = 1.0 x 10^-7). Correct answer is 10.30. Please show work to show why this is the correct answer. Thank you!

Answer 1

S032- + H20 ---> HS03- + OH-

using ICE table

initial conc of S032- , HS03- , OH- are 0.4 , 0 , 0

change in conc of S032- , HS03- , OH- are -y , +y , +y

equilibrium conc of S032- , HS03- , OH- are 0.4 - y , y , y

now

K2 = [HS03-] [OH-] / [S032-]

1 x 10-7 = [y] [y] / [0.4-y]

y2 + ( 10-7) y - ( 4 x 10-8) = 0

solving we get

y = 2 x 10-4

now

[OH-] = y = 2 x 10-4

we know that

pOH = -log [OH-]

pOH = -log 2 x 10-4

pOH = 3.7

now

pH = 14 - pOH

pH = 14 - 3.7

pH = 10.3

so

pH of 0.4 M Na2S03 solution is 10.3