Question

Calculate the pH of 0.15 M Co(NO₃)₂. For [Co(OH₂)₆]²⁺, K_a = 5.0 x 10^-10.

PLEASE SHOW ALL WORK.

Answer 1

Co(NO3)2 dissociates in water to give Co2+ ions

We have equation

Co2+ (aq) + 6H2O (l) ---> [Co(H2O)6]2+

now acid dissociation eq is given by

[Co(H2O)6]2+ (aq)   <---> [Co(H2O)5OH-]2+ (aq) + H+ (aq)

at equilibrium [ Co(H2O)6]2+] = 0.15-X ,   [ Co(H2O)5OH-] 2+ = [OH-] = X

Ka = [ Co(H2O05OH-]2+] [H+] / [ Co(H2O)6]2+

5 x10^-10 = X^2 / ( 0.15-X)              ( approximating 0.15-X nearly equal to 0.15 since Ka is very small we get very low value of X)

X= 8.66 x 10^ -6 = [H+]

pH = -log [H+]

     = -log ( 8.66x10^-6)

    = 5