kA and kB

[Nigah Fatima]

I am still confused on what exactly kA and kb are, and then for that case I am not sure what it means when we take the pH of these values (pkA and pkB). I understand that kA and kB values are for weak acids only, but then I saw some textbook problems where we calculated for strong bases. If someone can also explain with a calculation example, that would be very helpful.

[Diego Gonzalez 1H]

Ka is the acid dissociation constant and it is a measure of the strength of an acid in solution. On the other hand, Kb is the base dissociation constant and it is a measure of the strength of a base in a solution. pKa is used to compare the strength of acids while pKb is used to compare the strength of bases. pKa is the -log of the acid dissociation (Ka). Similarly, pKb is the -log of Kb.

An example calculation. If we take the Ka of HCN (6.2 x 10^-10) and we take the -log of this, we would get the pKa, 9.2.

[Ashley Wilson 2L]

Ka measures the strength of a weak acid and Kb measures the strength of a weak base. Since Ka is found using the concentration of the products divided by the concentration of the reactants, a larger Ka value means the weak acid dissociated more.

[Anthony_Rio_3K]

Ka is the disassociation constant and a higher Ka value corresponds to a strong acid.

[Shawn Suen]

Ka is the equilibrium constant for an acid, Kb is the equilibrium constant for a base.

[Lawrence Javelo Disc 3B]

kA and kB are the equilibrium constants in the equations of acid and base reactions. Higher kA values mean that the acid is strong (because there are more products, those being hydronium) and high kB values mean that the base is strong.

[LillianHarvey2C]

Thank you guys, this topic makes a lot more sense to me now:)

[Ashley Fakolujo 3H]

I’ve seen how to calculate Ka , but not not kB. How would you go about doing this?

[chemoyku dis 1B]

I’ve seen how to calculate Ka , but not not kB. How would you go about doing this?

Hi!

It has the same logic behind it.

If we represent the reaction with a base and water (base dissociating in water to become aqueous):

B + H20 = (double headed arrows) BH+ + OH-

Since we learnt KA = products/ reactants, it would be the same for Kb, where you would put the concentration of the products/ reactants

Example: [BH+] [OH-]/ [B]

(we dont include water because its concentration is really large and constant)

I also would not worry about this so much currently because I believe Dr. Lavelle would go in depth about how to calculate these in 14B!

[Ashley Fakolujo 3H]

Thank you so much!