If the ka of a monoprotic weak acid is 3.1 × 10-6, what is the ph of a 0.14 m solution of this acid?

PH is calculated by using following formula,

                              pH  =  - log [H⁺]

As we are not given with [H⁺], so we will calculate it bu using following formula,

                 [H⁺]  =  

Putting Values and solving,

                 [H⁺]  =  

                 [H⁺]  =  

                 [H⁺]  =  4.70 × 10⁻⁴

Now,
                 pH  =  - log (4.70 × 10⁻⁴)

                 pH  =  3.32

What? i don’t understand

Sorry i dont understand the question can you rephase it?