Asked by Angela
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
Answered by
DrBob222
...........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^+).
Answered by
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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