Question

When 2.3 g Co(NO3)2 is dissolved in 0.375 L of 0.44 M KOH, what are [Co2+], [Co(OH)42- ], and [OH- ]??? (Kf of Co(OH)42- = 5.0 109)

Answer 1

Co(NO3)2 moles = mass/molar mass = 2.3/182.94 = 0.0125724

Co2+ moles = Co(NO3)2 moles = 0.0125724

KOH moles = M x V = 0.44 x 0.375 = 0.165 = OH- moles

Co2+ (aq) + 4OH- (aq) <---> [Co(OH)4]2- ,

OH- moles required as per 0.0125724 Co2+ moles = 4 x 0.0125724 = 0.05 , but we have 0.165 moles OH-

hence we have excess KOH and Co2+ is limiting reagent.

Hence [Co(OH)4]2- moles = Co2+ moles = 0.0125724

excess OH- moles = 0.165-0.05 = 0.115

now we have reverse equilibrium

[Co(OH)4]2-(aq) <---> Co2+ (aq) + 4OH- (aq)

now at equilibrium [Co(OH)4]2- = ( 0.0125724-X) /0.375 = 0.0125724/0.375 = 0.0335 since we get X very low value

0.0125724-X is nearly equal to 0.0125724

, [Co2+] = X/0.375 ,

[OH-] = (0.115+X)/0.375 = 0.115/0.375   = 0.3067   ( since we get X very low value 0.115+X nearly equal to 0.115)

1/Kf = [Co2+] [OH-]^4 /[Co(OH)4]2-

1/(5x10^-9) = [Co2+] ( 0.3067^4) / (0.0335)

[Co2+] = 7.6 x 10^ -10 M

[OH-] = 0.3067 M

[Co(OH)4]2- = 0.0335 M

X