Enough monoprotic acid is dissolved in water to generate 0.0163 M solution. The pH of the resulting solution is 6.31. Calculate the Ka of the acid.
My attempt at a solution:
[H+]=10^(-6.31)=4.898*10^-7
[A-]=[H+]=4.898*10^-7
[HA]=0.0163-4.898*10^-7=0.0162995
Ka=((4.898*10^-7)(4.898*10^-7))/(0.0162995)
Ka=1.47*10^-11
apparently what I calculated for Ka is incorrect, the hint says that I treated the initial H+ concentration as 0, the initial 10^-7 M H+ from water is not negligible. If [H+] initial=10^-7 and [H+] final =4.9*10^-7 by how much does the concentration change? What does that say about how much A- was produced?