Molarity of acid=2.5M
pH=5.1.
ka=?
Now
We need to write an eqn to show the dissociation of the acid
HA + H2O === H3O+ + A-
Writing The Equilibrium(Or Acid dissociation constant) of this reaction
Ka =[H3O+] {A-]/ {HA].
The concept behind this is
concentration of Products divided by those of reactants. Water is not written because its a pure liquid and does not affect the Equilibrium constant.
Now If you have any Idea on ICE tables..
You'd know that the concentration of acid will decrease by 2.5-x
Whilst the products...will increase by x each
Note: This is when the ratio of their Moles are in 1:1
ka= x.x/2.5-x
Since the Moles of A- and H3O+ are in 1:1... Their concentrations at equilibrium will be the same
so
Ka= x²/2.5-x
Now what is x??
x is the Hydrozonium ion concentration.
we can get it from the pH formula
pH= -log (H3O+)
Making H3O+ subject by applying Logarithm Rules
H3O+ = 10^-ph
x=10^-5.1
=7.94x10^-6.
Now back to Ka
Ka= x²/2.5-x
Ka= (7.94x10^-6)²/2.5-(7.94x10^-6)
Ka= (7.94x10^-6)²/2.4999
Ka= 2.52x10^-11.
Was a Fun One
