...........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^+).
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
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
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
1 answer