Question

Calculate the pH of an aqueous solution of 0.234 M potassium sulfite. pH- Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

Potassium sulfite in K2SO3 It is a wrong erecevolyte and it dimeciates completely in water to art and so 2 - Soz is a conjugate base of weak and H2SO3- So accepte evera Proton from water 50.3 + H2O 2 H50g + Oh Khan = 0.71418 1505 + H2O ² H2SO3 + oH Killz = grisextoo i Kby = 0.159x106 Kb2 = 0.714X1022 Given that Potassium sulfite il 0.234 m 3 [so}-Vinitial = 0.234 m

So} + H2O 2 HSO3 + oH I 0.234 c -X +x. ta th E 0.234-2 Kbi = [1sozlequillibrium Cow Jewillibrium 0.234-2 0.159x105 - * 0.234-X. Neglecting an in the denominator 22- 0.234 x0.159 x156 = 0.03721x106 x = 0.193 x103 m. [1503] ==0-193 xiôm - [oH) = x = 0.193x10²m

H₂O ² HSO3 + 0.193x103 H2SO3 + O OH- 0.193x163 I c ta + x -X [ 0.193x10²x - - 0.193x183+a Neglect (0.193x16? Kb = considering in the terme (0.193 x163-) I neglect a is small, we can neglect it shown by arrow marks above Kb = 1.0.193x103 0.193x163 :* = 0.714 xiót? m ... [OH-] = 0.214 x162 m

1. Total [om-] = (0.193x163 + 0.14 x1õl2 ) m = 0.193x173 m pole = -log(OH-] = -log(0-193x163 ) = 3-714 pH = 14 - pOH 14-3.714 = 10-286 Question 2 whether it is potanium sulfite or sodium sulfite, the pH = 10.286 because initial conc. of sulfite son a same Na₂SO3 is also strong electrolyte like K₂SO3 - Therefore (50,3 Timikal = conc. of Nazdo3 70.234 m - pH = 10.286 :