Remember,
pH=-log[H^+]
so,
pH=-log[x]
I made a bobo, which may lead to some confusion.
Here is my post revised:
You don't need it. Just calculate the new molarity of the acid after addition of NaOH, and then determine the molarity of the acid in solution.
How many moles of propionic acid are left after addition, of the strong base?
In this case, Molarity=moles
1.29 x 10^-2 M=1.29 x 10^-2 moles
1.29 x 10^-2 moles-3.87 x-3.87 x 10^-3 moles=9.03 x 10^-3 moles of propionic acid.
9.03 x 10^-3 moles of propionic acid=9.03 x 10^-3 M of propionic acid
HA -----> H^+ + A^-
ICE Chart is the following:
................HA ................H^+................A^-
I.....9.03 x 10^-3 M.........0........3.87x 10^-3 M
C.............-x....................x....................x
E..9.03 x 10^-3 M-x........x.....3.87x 10^-3 M+x
Ka=1.3×10-5=[H^+][A^-]/[HA]
or
Ka=[x][x+3.87 x 10^-3 M ]/[9.03 x 10^-3 M-x]
After plugging in the E line into the equation.
Assuming dissociation is small, the equation becomes:
Ka=[x][3.87E-3 M ]/[9.03 x 10^-3 M]
Solving for x:
x=Sqrt*{[Ka*9.03 x 10^-3 M]/3.87E-3 M}
Check 5% dissociation.
1 answer
1 answer