At equilibrium, what is the hydrogen-ion concentration if the acid dissociation constant is 0.000 001 and the acid concentration is 0.01M?

can you help me step by step and explain?

2 answers

...........HA --> H^+ + A^-
I.........0.01....0......0
C..........-x.....x......x
E........0.01-x...x......x

Ka = 1E-6 = (H^+)(A^-)/(HA)
Substitute the E line into Ka expression and solve for x = (H^+).
acid(aq)>>H+ ??

keq=[H+)[??]/(.01-x)

1E-6=x^2/(.01-x)

x^2+xE-6-1E-8=0

x= (-E-6+-(1E-12 +4E-8)/2

x= you work it out, about
= -.5E-6+1E-4 about .000095