Question
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?
can you help me step by step and explain?
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^+).
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^+).
bobpursley
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
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