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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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    I don't get that answer. First, the (H2O) appearing in the Ka expression is incorrect. That number (usually close to 55.5 M) has been included in the Ka you find in the tables. The spacing didn't show up right for the ICE table so I can't say if they are right or wrong but I CAN see that you have included a number for (H2O) which should not have been used. Third, You used 1.33 for the molarity of the HCN but the problem states 1.24 M. The final equation which have is
    (H^+)(CN^-)/(HCN)= Ka
    (x)(x)/(1.24-x) = 6.2 x 10^-10
    x = (H^+) = 2.77 x 10^-5 and pH = 4.56.
    Check my work. Let me know if there is anything you don't understand.

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