Hello!
The dissociation reaction of a monoprotic weak acid is the following:
HA + H₂O ⇄ A⁻ + H₃O⁺
The ka value is expressed as follows:
[tex]Ka= \frac{[A^{-}]*[H_3O^{+}] }{[HA]} [/tex]
After the dissociation, we can clear for the concentration of H₃O⁺ (x) in the following way (Considering that x is small).
[tex]Ka= \frac{x*x}{[HA]_i-x} \\ \\ x= \sqrt{[HA]_i*Ka}= \sqrt{0,18M*5,1*10^{-6} } \\ \\ x=0,000958M=[H_3O^{+} ][/tex]
So, to finish, we apply the definition of pH
[tex]pH=-log[H_3O^{+}] =-log (0,000958M)=3,02[/tex]
So the pH of this solution is 3,02
Have a nice day!
