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.
Remember,
pH=-log[H^+]
so,
pH=-log[x]
To find the number of moles of propionic acid left after the addition of the strong base, you can follow these steps:
1. Convert the given molarity of the acid, 1.29 x 10^-2 M, to moles by multiplying it by the volume of the solution in liters. Let's assume the volume of the solution is 1 liter for simplicity:
moles of propionic acid = 1.29 x 10^-2 M * 1 L = 1.29 x 10^-2 moles
2. Subtract the moles of NaOH added, which is 3.87 x 10^-3 moles, from the initial moles of propionic acid:
moles of propionic acid left = 1.29 x 10^-2 moles - 3.87 x 10^-3 moles = 9.03 x 10^-3 moles
So, there are 9.03 x 10^-3 moles of propionic acid left after the addition of the strong base.
Alternatively, you can calculate the molarity of the propionic acid left in solution. Since molarity is equal to moles divided by volume, in this case, both the moles and volume are the same, which is 1 liter. Therefore, the molarity of the propionic acid left is also 9.03 x 10^-3 M.
The given information indicates an ICE (Initial, Change, Equilibrium) chart for the reaction of propionic acid, HA, dissociating into H+ and A-. This chart helps in determining the concentrations of the species at equilibrium.
To calculate the concentration of each species at equilibrium using the ICE chart, you can follow these steps:
1. Initially, the concentration of HA is 9.03 x 10^-3 M, and the concentrations of H+ and A- are both zero.
2. The change in concentration for the H+ species will be x, and for the HA and A- species, it will be -x (equal magnitude but opposite sign).
3. Therefore, at equilibrium, the concentration of HA will be 9.03 x 10^-3 M - x, the concentration of H+ will be x, and the concentration of A- will be 3.87 x 10^-3 M + x.
Now, the given Ka expression is used to find the value of x, assuming dissociation is small:
Ka = [H+][A-] / [HA]
Substituting the concentrations from the equilibrium line of the ICE chart into the Ka expression:
Ka = x * (3.87 x 10^-3 M + x) / (9.03 x 10^-3 M - x)
If the dissociation is small, the concentration of x is negligible compared to the concentrations in the equation. So you can simplify the equation further:
Ka ≈ x * (3.87 x 10^-3 M) / (9.03 x 10^-3 M)
Ka ≈ (x * 3.87 x 10^-3 M) / (9.03 x 10^-3 M)
Now you can solve for x by rearranging the equation and considering that the Ka value is known:
x ≈ √[(Ka * 9.03 x 10^-3 M) / (3.87 x 10^-3 M)]
This calculation will give you an approximation of the value of x.
Finally, you mention "Check 5% dissociation." This might refer to determining whether the calculated value of x is less than 5% of the initial concentration of propionic acid (1.29 x 10^-2 M). If the value of x is less than 5% of the initial concentration, then the assumption of small dissociation is valid. You can compare the value of x to 5% of 1.29 x 10^-2 M to check for this condition.