Question: Enough monoprotic acid is dissolved in water to generate 0.0163 M solu

Question

Question:
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?

Explanation / Answer

H+]= 10^-6.31 =6.1659*10^-7 M HA H+ + A- start 0.0163 change -x .. . . +x . . .+x at equilibrium 0.0163-x. . . x. . .x x = 6.1659500*10^-7 M= [H+]= [A-] [HA]= 0.0163 - x=0.01629 M Ka = [H+][A-]/ [HA]= ( 6.16559500*10^-7)(6.16559500*10^-7)/ 0.01629=2.333*10^-11

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