why is it harder to determine experimentally the pka and ka of Hcl?

Question

in class, we experimentally recorded pH vs volume of base added to find the equivalence pt and therefore, pka. a question asks why we didnt experimentally derive a strong base/strong acid (HCl). when i googled HCl's pKa, it is a negative number, so finding it graphically would be difficult. is that why ?

Answer 1

Hello, Lily,

pKa is the is the negative log of the acid dissociation constant or Ka value. A lower pKa value indicates a stronger acid, such as HCl. I'm actually not sure why a pKa would be a negative value. Please check that.

K_a for HCl should be 10⁶ . The equilibrium and equilibrium constant can be written as the following:

HCl + H₂O = H₃O⁺+ Cl^-

K_a = [H₃O⁺][Cl^-]/[HCl]

HCl is a strong acid and thus nearly completely dissociates in water. The K_a is calculated in a manner that the equilibrium concentration of undissociated HCl is in the denominator. So I'm predicting that it might be more difficult to get good numbers, owing to the fact that you are dividing by a very small number, and any deviation makes a relatively large impact on the result.

If we start with 1 M HCl, consider what happens in the equilibrium equation. The concentrations of both the [H₃O⁺] and the [Cl^-] are equal since we get equimolar amounts of both as per the balanced equation. Let's say that x moles of HCl dissociate. That means we'll get x moles of both H₃O⁺ and Cl^- . The remaining concentration of HCl is thus (1M - x). If K_a is 10⁶, we can see that x must be a large number approaching 1M. If x = 0.9, which, for a starting concentration of 1M, means 90% of the acid dissociated, we get a calculated equilibrium of:

(0.9)²/(1-0.9) = 8.1

That's a LONG way from 10⁶.

Even if we go to 99.9990 percent dissociation, we only get up to 1x10^4.

We have to go to 99.9999 percent to reach 10^-6.

My conclusion is that measurements requiring this level of precision are likely to be very difficult.

Bob