You will learn how to mathematically manipulate the values of Ka, Kb, pH, and measured concentrations of hydroxide and hydronium ions.
For the hypothetical acid, HA, we can write
HA(aq) + H2O(l) --> H3O+(aq) + A-(aq)
The equilibrium constant is then (remember that the concentrations are equilibrium concentrations)
Ka = [H3O+][A-]/[HA]
But the concentrations of both products will be equal at equilibrium because they each have the same stoichiometric coefficient, so we have
Ka = [H3O+]2/[HA]
We then see the relationship between Ka and the equilibrium values of [H3O+], or pH, and [HA]. Let us now use experimental data and the above equation, or its base analog, to calculate the unknown quantity.
What is the pH of a 1.60 M solution of HCN? We first look up the Ka of HCN and see that it is 4.0 x 10-10. Now let's set up a working table of data as we have done before.
|Equilibrium(M)||1.60 - x||x||x|
So we have that Ka = 4.0 x 10-10 = (x)(x)/(1.60 - x) = x2/(1.60 - x). We can solve this quadratic equation or we can assume that x is much less than 1.60 which gives the simple equation x2 = 1.60 (4.0 x 10-10) = 6.4 x 10-10 or x = 2.53 x 10-5. This x is the equilibrium value of the hydronium ion concentration so the negative log of this number will be the pH. Here the pH is 4.60. If you solve the quadratic equation instead of making the simplifying assumption you will obtain the same answer to three significant figures. This approximation is usually valid if the initial concentration of acid is greater than 100 Ka.
The other variations on this theme are rather straightforward and can be practiced with the review problems below. We can also calculate the percent ionized by dividing the quantity of acid ionized by the initial acid concentration and multiplying by 100.
Web Author: Dr. Leon L. Combs
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