Question

Cr3+ (and other metal ions with a charge ≥ 2) is acidic by virtue of the hydrolysis reaction. Cr3+ + H2O equilibrium reaction arrow Cr(OH)2+ + H+ Ka1 = 10−3.66 [Further reactions produce Cr(OH)2+, Cr(OH)3, and Cr(OH)4−.]

a) Considering only the Ka1 reaction, find the pH of 0.019 M Cr(ClO4)3.

b) What fraction of chromium is in the form Cr(OH)²⁺?

Answer 1

Given

Ka1 = 10^-3.66 = 2.18776 x 10-4

now

consider the hydrolysis reaction

Cr+3 + H20 --> Cr(OH)+2 + H+

using ICE table

initial conc of Cr+3 , Cr(OH)+2 , H+ are 0.019 , 0 ,, 0

change in conc of Cr+3,Cr(OH)+2 ,H+ are -y , +y , +y

equilibrium conc of Cr+3,Cr(OH)+2,H+ are 0.019-y , y, y

now

Ka1 = [Cr(OH)+2] [H+] / [Cr+3]

2.187776 x 10-4 = [y] [y] / [0.019-y]

y2 = 2.18776 x 10-4 x (0.019-y)

y2 + ( 2.18776 x 10-4)y - ( 4,1562 x 10-6) = 0

y = 1.9 x 10-3

so

[H+] = y = 1.9 x 10-3

we know that

pH = -log [H+]

pH = -log 1.9 x 10-3

pH = 2.721

so

pH of the solution is 2.721

2)

now

fraction of Chromium as Cr(OH)+2 = [Cr(OH)+2] / initial concentration of Cr(Cl04)3

now

[Cr(OH)+2] = y = 1.9 x 10-3

so

fraction of Chromium as Cr(OH)+2 = 1.9 x 10-3 / 0.019

fraction of Chromium as Cr(OH)+2 = 0.1