Question

The acid ionization constant for Co(H2O)62+(aq) is 3.0×10-10. Calculate the pH of a 0.0229 M solution of Co(NO3)2.

Answer 1

The Co(NO₃)₂ will dissociate in Co + 2 NO₃, the Co ions will have a concentration of 0.0229 M

Then the Co will react with water to form

..................Co⁺² + 6H₂O (l) ---> [Co(H₂O)₆]²⁺

..................0.0229.........................0.0229

The complex will dissociate like :

..........[Co(H₂O)₆]²⁺ =========== [Co(H₂O)5OH-]²⁺ + H⁺; let´s write the ICE chart

I...........0.0229.......................................0............................0

C............-x............................................x..............................x

E.......0.0229-x....................................x...............................x

as you can see the concentration of H is equal to x

remember that the K equilibrium is expressed as the relationship between products and reactants

K = products / Reactants so for our particular case

K = [x] * [x] / [0.0229 - x ] = 3 x 10^-10

x² / (0.0229 - x) = 3 x 10^-10

let´s simplify this, since Ka is very small we can also expect a very small value for the x so, 0.0229 - x might be reduced to just 0.0229 so:

x² / (0.0229) = 3x10^-10

x² = 0.0229 * 3x10^-10 = 6.87 x10^-12

x = 2.621 x 10^-6 M , this is the concentration of H

PH = - log (H)

PH = -log ( 2.621 x 10^-6 ) = 5.58

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