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April 21, 2014

Homework Help: chemistry

Posted by Amandeep on Sunday, April 20, 2008 at 2:08pm.

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

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