rachel liu 3k wrote:
Julie_Reyes1G wrote:For the scope of this class, KA is just a measure of the relative strength of an acid. The higher the KA, the stronger the acid. The pKA is the -log KA. The lower the pKA, the stronger the acid. There is also KB and pKB (the same definitions but for a base instead. Usually, you will be given KA and asked to solve for pKA, or vice versa. You would use the equation above.
pH is a measure of whether something is an acid or a base. For this class, anything pH<7 is acidic, pH>7 is basic, and pH=7 is neutral (values will fall between 0 and 14). In terms of calculation, pH is the -log of the concentration of H+ ions. You can also calculate pH if you are given pOH, with the equation pH+pOH=14. If you are given concentration of OH- ions, you can find pOH and solve for pH as well.
pH and pKA are not directly related through calculations. But, something with a low pH (indicates a strongish weak acid) will most likely have a low pKA too.
Also do you remember what professor was saying about how there is not pKa or Ka or a strong acid or pKb or Kb of a strong base because it will be infinity? why is that?
measures the extent to which an acid dissociates, and since a strong acid dissociates completely/infinitely, the value is basically infinity. If you want to know why, I can explain in more depth below (but for all intents and purposes, explaining it that way is enough for the final).
I'm going to give you kind of a math-y explanation, so bear with me lol. This has to to with the definition of strong acids/bases. In theory, strong acids and bases dissociate completely in water. So the strong acid HCl, for example, will dissociate completely into H+
in water. There will be no HCl molecules left.
=[H+][A-]/[HA]. So for HCl it would be KA
=[H+][Cl-]/[HCl]. But as I said, since the acid completely dissociates, there is no concentration of HCl left. So [HCl] would equal 0. If we substitute that in, we get KA
=[H+][Cl-]/0. Anything with a denominator of 0 approaches infinity. That's the theory behind it, but probably something Dr. Lavelle will explain more in Chem 14B.