The Ka for HCN is 6.2 x10^-10. What is the pH of a 1.00 M solution of LiCN?

This is what I came up with....

given...Ka=6.2 * 10^-10

LiCN + H2O <==> HCN +LiOH

1.00M X X

These x's are supposed to be under HCN and LiOH respectively the system puts the x's together for some reason

Solve for X

(6.2*10^-10) = x^2/1.00M

x=2.49*10^-5= [OH]

Find pOH= -log of the above number

then 14-pOH=pH= 9.40

I don't know if I'm right but given the info you gave me this is how you calculate pH

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The Ka for HCN is 6.2 x10^-10. What is the pH of a 1.00 M solution of LiCN?

I was thinking that HCN is a weak acid and LiCN is a weak base so the Kb of the CN- in LiCN would be the same as the Ka of HCN. But that seems wrong.

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Is this compound supposed to be NaCN or HCN? Assuming that it is HCN (since NaCN does not dissociate a hydronium ion), we see that HCN will dissociate as follows: HCN(aq) ⇆ H+(aq) + CN-(aq). The base-acid expression for this reaction is: Ka = ([H+] * [CN-])/[HCN]. Since the [H+] and [CN-] ions are produced equally (seen from the coefficients of the balanced chemical equation), we can let x = [H+] = [CN-]. With [HCN] = 0.021 and Ka = 4.9 x 10^-10, we have: 4.9 x 10^-10 = x^2/0.24 ==> x = √(0.24 x 4.9 x 10^-10) = 1.08 x 10^-5. This gives [H+] = 1.08 x 10^-5 and: pH = -log[H+] = -log(1.08 x 10^-5) = 4.97. So the answer is the none of the above. I hope this helps!

Ka Of Hcn