Question

If the ka of a monoprotic weak acid is 5.1 × 10-6, what is the ph of a 0.18 m solution of this acid?

Answer 1

Hello!

The dissociation reaction of a monoprotic weak acid is the following:

HA + H₂O ⇄ A⁻ + H₃O⁺

The ka value is expressed as follows:


`Ka= ([A^(-)]*[H_3O^(+)] )/([HA])`

After the dissociation, we can clear for the concentration of H₃O⁺ (x) in the following way (Considering that x is small).


`Ka= (x*x)/([HA]_i-x) \\ \\ x= √([HA]_i*Ka)= \sqrt{0,18M*5,1*10^(-6) } \\ \\ x=0,000958M=[H_3O^(+) ]`

So, to finish, we apply the definition of pH


`pH=-log[H_3O^(+)] =-log (0,000958M)=3,02`

So the pH of this solution is 3,02

Have a nice day!