K value of strong acids and bases

[Aanya Pramanik 1C]

Why is there no Ka value for strong acids and no Kb value for strong bases?

[Arad_Badiee_2B]

The reason for this is that with a strong acid, since it dissociates completely, if we were to have a value of zero in our denominator for the dissociation constant, then our dissociation constant value would be undefined.

[Joshua Lee 3C]

The general equation for Ka is [H+][A-]/[HA] where [HA] represents the concentration of the original acid, [H+] represents the concentration of the hydrogen ions produced, and [A-] represents the concentration of the conjugate base produced. However, with strong acids, we assume that the acid 100 percent dissociates or at least gets very close to it. Thus the [HA] value of the equation would be very very small or even 0. If this were the case our Ka value would be very large and because of that we do not calculate or pay attention to this value. We just assume that it is a very large number. It is a similar case with strong bases.

[Samantha Melin 2E]

A strong acid would have a concentration of close to zero. Because this value would be inserted in the denominator of our equation to calculate Ka, it would produce a very large value for Ka. Knowing this, we can just assume that our Ka value is a very large number. The same goes for calculating Kb in a strong base. The number we are inserting into the Kb equation denominator would be very small, resulting in an extremely large Kb value for a strong base.

[LukeYing_3H]

A strong acid would have a concentration close to 0, which would result in a very large Ka value due to the equation putting this value in the denominator. The same is true for Kb.

[Molly McAndrew 1 1H]

So, do we need to know/apply this formula, or can we just assume we're working with only strong acids and bases?

[Tony Chen 1F]

I think Dr. Lavelle mentioned we will mostly be working with strong acids and bases for 14A. But on tests it should be specified what we're working with

[Cory Poon 3G]

To calculate Ka, you divide the concentration of [A-][H+] by [AH], and since usually all of AH dissociates, the dissociation constant will be a VERY large number, so assume that AH all dissociates, making AH very small, and makes the Ka number huge.