Question

In a 1.1400 M aqueous solution of a monoprotic acid is ionized. What is the value of Ka?

I did

HA <=> [H+] + [A-]

I 1.14 o o

C -x +x +x

E 1.14-x x x

Ka= [H+][A-]\[Ha]

=x^2/1.14-x   

and I have no idea where to go from here

Answer 1

HA <=> [H+] + [A-]

I 1.14 o o

C -x +x +x

E 1.14-x x x

Ka= [H+][A-]\[Ha]

     =x^2/1.14-x                    x<<<1.14

Ka = x²/1.14

in this proble percentage of acid is 3.82%    x = 3.82*1.14/100   = 0.0435

    =x^2/1.14-x

    = 0.0435*0.0435/1.14-0.0435

    = 0.00189/1.0965   = 0.00172

Ka = 1.72*10^-3

correct problem

In a 1.1400 M aqueous solution of a monoprotic acid 3.82% of the is ionized. What is the value of Ka?

it is missing in the problem