Asked by Jason
What is the original molarity of a solution of ammonia whose pH is 11.11
okay i tried to convert to pOH- so i got 2.89 and then to [OH]=1.288e-3
and then solve using Kb=1.8e-5
whats going wrong???
okay i tried to convert to pOH- so i got 2.89 and then to [OH]=1.288e-3
and then solve using Kb=1.8e-5
whats going wrong???
Answers
Answered by
DrBob222
The pOH is correct as is the (OH)^-. Show your work for the rest of it.
I would do
[(1.288 x 10^-3)^2/(y-1.288 x 10^-3)] = 1.8 x 10^-5 and solve for y. I get 0.0935 M but check my work. Without the correction for y-0.001288 (that is, calling it just y), I get 0.0922 M
I would do
[(1.288 x 10^-3)^2/(y-1.288 x 10^-3)] = 1.8 x 10^-5 and solve for y. I get 0.0935 M but check my work. Without the correction for y-0.001288 (that is, calling it just y), I get 0.0922 M
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