Asked by Amandeep
What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10-10?
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For your understanding:
HCN ⇌ H+ + CN-,
Ka = 6.2x10^-10 = [H+]*[CN-]/[HCN]
Notice the H+ and CN- ions are generated in pairs during HCN dissociation, hence [H+] = [CN-]. Hence:
[H+]^2 = 6.2x10^-10*[HCN] = 7.688x10^-10
pH = -log([H+]) = -0.5*log(7.688x10^-10) = 4.56
Rate my answer please
For your understanding:
HCN ⇌ H+ + CN-,
Ka = 6.2x10^-10 = [H+]*[CN-]/[HCN]
Notice the H+ and CN- ions are generated in pairs during HCN dissociation, hence [H+] = [CN-]. Hence:
[H+]^2 = 6.2x10^-10*[HCN] = 7.688x10^-10
pH = -log([H+]) = -0.5*log(7.688x10^-10) = 4.56
Answers
Answered by
DrBob222
Two things.
1st. It would help for understanding if you showed next to last step as
[H^+]^2= 6.2 x 10^-10*1.24 = 7.688 x 10^-10.
2nd. I'm not sure how clear the last step is. I prefer to take the square root of 6.2 x 10^-10 = ??, THEN do the pH = -log(H^+). The way you have done it the reader must recognize that you took the log of the squared term so you must divide by 2.
1st. It would help for understanding if you showed next to last step as
[H^+]^2= 6.2 x 10^-10*1.24 = 7.688 x 10^-10.
2nd. I'm not sure how clear the last step is. I prefer to take the square root of 6.2 x 10^-10 = ??, THEN do the pH = -log(H^+). The way you have done it the reader must recognize that you took the log of the squared term so you must divide by 2.
Answered by
aaa
4.95
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