A 2.5 M solution of a weak acid is prepared. Using a pH meter, the pH is measured to be 5.1. Calculate the acid ionization constant, , of this weak acid. Show your work

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