Ka of Strong Acids

[michaelcrisera]

In the lecture, Dr. Lavelle mentioned that strong acids are not assigned Ka values. Can someone explain why this is?

[Alex Luong 3H]

We assume that strong acids (and bases) dissolve completely. When using the formula to find the Ka of a strong acid, the denominator would be 0, because it completely dissolves, and the result would be undefined.

[Stephen_Kim_1D]

Strong acids completely dissociate so the Ka, with a denominator of 0 or a very very small number, is either undefined or a very very large number (and it doesn't make sense to assign either of these values to the strong acid).

[Sidharth Paparaju 3B]

Image the Ka value as = [H+][Conjugate base]/[acid]

In a strong acid, theoretically all of the acid is disassociated, which means none is left. Therefore, the Ka would have a denominator of zero, which would mean that the Ka would theoretically be infinity.

[Reagan Feldman 1D]

We assume strong acids (and bases) to be very large values, but they don't have specific Ka values. In terms of strong acids, the acid would be 100% dissociated, therefore when calculating the Ka, the denominator (A-H concentration) would be zero, hence we would get an extremely large/undefined value.

[August Blum Dis 3D]

Since they dissociate completely, there is no need for them to have Ka values. Calculating their Ka would involve dividing by 0, and thus could not be done without getting an undefined value.

[Anish_Marripati_2F]

Since strong acids dissociate completely, they don't have a ka value.

[Divya Mehta 2K]

Strong acids dissociate completely, therefore there is no Ka value and the denominator is close to zero

[Sanskriti Balaji 1J]

Hi! There is no Ka value for strong acids or bases because they disassociate completely. Even if you were to calculate the Ka value, you would have to divide by 0, which would result in you getting an undefined value. Hope this helps!

[Aditya Desai 1A]

Ka would be undefined.

[Arjun_Anumula_3E]

Strong acids are not given a Ka because they are assumed to almost fully dissociate in water. The Ka values would be so large (since the equilibrium would favor the forward reaction) that it would not be productive to compare the Ka's among strong acids. However, weak acids often favor the reverse reaction, so it would make sense to compare the Ka's.