acid base question

[Captain Sisko]

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From a GS exam-

Cyanide ions (CN-) could also have been used to precipitate the cations. If after all the cations have been precipitated, the concentration of CN- is 0.02 M, calculate the pH of the solution given that:

CN- + H2O --> HCN + OH-
Kb = 1.39 x 10-5

a 4.9
b 5.4
c 7.7
d 10.7

I sort of get how they solved it, by using the definition of Kb as a function of concentrations to find pOH and then pH, but in so doing they make the assumption that [HCN] = [OH]. Can someone tell me why this is a valid assumption?

thanks all

for those who are using this as practice for themselves, the correct answer is D

 

[discowisco]

Hey so the way i did this was since kb = 1.39 x 10^-5 that means pkb is about 4.8 which makes pka about 9.2 thus ph = pka + log conj base/ conj acid.

I got stuck at that part, anyone know what the conj base and acid values would be?

 

[ChemTutor]

[HCN] = [OH-] follows from the stoichiometry of the reaction.

The only other source of OH- is due to autoionization of water. We ignore this source unless the pH we are getting as our answer is very close to 7.

 

[discowisco]

Is my method correct? How would you solve it with henderson hasselback?

 

[chem4ever]

Is my method correct? How would you solve it with henderson hasselback?

You can't use Henderson-Haaselbach here because we don't know the final [CN-] nor do we know [HCN]. You need to use the definition of Kb and solve for [OH-] like OP said.

And the previous poster is right: we can ignore previous OH- so any OH- in final solution would be due to reaction so for every molecule of HCN that is made, one molecule of OH- will be made as well.

 

[Captain Sisko]

[HCN] = [OH-] follows from the stoichiometry of the reaction.

The only other source of OH- is due to autoionization of water. We ignore this source unless the pH we are getting as our answer is very close to 7.

thanks pal, that makes sense.

 

[futureDPM27]

Is the answer 10.7?