Asked by Ab
What is the PH of a 0.747M sodium cyanide solution (NaCN) knowing that the Ka of Hydrocyanic acid is 4.9x10-10?
Answers
Answered by
DrBob222
The pH is determined by the hydrolysis of the CN^- as follows:
.....................CN^- + HOH ==>HCN + OH^-
I..................0.747.......................0..........0
C...................-x...........................x...........x
E.................0.747-x.....................x...........x
Kb (for CN^-) = Kw/Ka(for HCN) = (HCN)(OH^-)/(CN^-)
Plug the E line into the Kb expression and solve for x = (OH^-). Convert to (H^+) and pH. Post your work if you get stuck.
.....................CN^- + HOH ==>HCN + OH^-
I..................0.747.......................0..........0
C...................-x...........................x...........x
E.................0.747-x.....................x...........x
Kb (for CN^-) = Kw/Ka(for HCN) = (HCN)(OH^-)/(CN^-)
Plug the E line into the Kb expression and solve for x = (OH^-). Convert to (H^+) and pH. Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.