Destroyer g.chem #33

[hunterpostbacst]

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Ok, I understan a bit and can't totally agree on the solution.

1) What if we can get Ka directly?
HCN + H2O = CN- + H3O+
The solution says CN- + H2O = HCN + OH- and got Kb and then Ka from Kb

2) I don't understand how M of NaCN can sneak into [CN-]

Anybody can explain better than orgoman?

Thanks!

 

[UCB05]

My understanding is that addition of the -CN anion creates a basic solution, so that it's easier to calculate [-OH] through a kb expression, since the rxn mechanics involve a cyanide base reacting with water to create its conjugate acid HCN and hydroxide. Then once you have [-OH] it's easy to figure out [H+]. Through the kb expression kb=[-OH][HCN]/[-CN] you can calculate by hand the concentrations of hydroxide and HCN with the assumption that ([-OH] = [-CN] = x) and (.040 - x ~~ .040), whereas with acid dissociation ka = [H+][-CN]/[HCN] you have two unknowns as well, but theyre in the numerator AND denominator, which fudges things up.

2) NaCN dissociates into its constituent ions, and in the case of -CN being the conjugate base of a weak acid, contributes to its formation through the common ion effect

 

[hunterpostbacst]

Thanks a lot!