February 27, 2017

Homework Help: chemistry

Posted by Abigail on Tuesday, April 15, 2008 at 5:01pm.

Please judge my answer, thank you

What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10-10?

It is imperative to initially set up ionization. THen, it becomes important to write the Ka expression, after which setting up an ICE table becomes nessecary. Substituting as well as solving for H^+ and converting this to pH becomes the next step. The ionization equation is HCN(aq) + H20==> H3O^+ CN^-. The Ka expression becomes Ka=[H3O^+][CN^-]/[HCN][H2O] After setting up the ice table the equilibrium concentration, in order, are x- 1.33 +x -x -x Solving for H^+ gives a value of 4,61. Converting H^+ concentraitons to pH requires the use of the formula pH=-log[H^+]. After substituting in the values, the answer becomes 0.7. Therefore the pH is 0.7

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