What is the percent ionization of a solution prepared by dissolving 0.0286 mol of chloroacetic acid in 1.60 L of water? For chloroacetic acid ,Ka=1.4x10^-3
Chloroacetic acid's formula is too long to write. Let's just call it HAcl.
(HAcl) = 0.0286mol/1.60L = about 0.018 but you do it more accurately.
........HAcl ==> H^+ + Acl^-
I.......0.018.....0......0
C........-x.......x.......x
E......0.018-x....x......x
Ka = (H^+)(Acl^-)/(HAcl)
Substitute Ka and the E line into Ka expression and solve for x = (H^+). Then
%ionization = (H^+)/(0.018)
Don't forget to redo the 0.018 number.
What you mean redo 0.018?
To find the percent ionization of a solution, you need to first determine the concentration of the dissociated ions and the initial concentration of the acid. Then, you can calculate the percent ionization using the formula:
Percent Ionization = (Concentration of dissociated ions/Initial concentration of acid) * 100
In this case, you are given the moles of chloroacetic acid and the volume of water. To calculate the concentration of chloroacetic acid, use the formula:
Concentration (in mol/L) = moles of solute/volume of solution (in L)
Given:
Moles of chloroacetic acid (CH2ClCOOH) = 0.0286 mol
Volume of water = 1.60 L
Concentration of chloroacetic acid = 0.0286 mol / 1.60 L
Once you find the concentration of chloroacetic acid, you can calculate the concentration of the dissociated ions. Chloroacetic acid (CH2ClCOOH) dissociates into a hydrogen ion (H+) and a chloroacetate ion (CH2ClCOO-). Since the acid dissociates in a 1:1 ratio, the concentration of H+ ions will also be equal to the concentration of CH2ClCOO- ions.
Now, let's calculate the concentration of hydrogen ions (H+):
[H+] = Concentration of chloroacetic acid (CH2ClCOOH)
The concentration of CH2ClCOO- ions will also be the same as the concentration of H+ ions.
Now, to calculate the percent ionization, we need the equilibrium expression for the dissociation of chloroacetic acid:
Ka = [H+][CH2ClCOO-] / [CH2ClCOOH]
Given:
Ka = 1.4 x 10^-3
[H+] = Concentration of chloroacetic acid (CH2ClCOOH)
[CH2ClCOO-] = Concentration of chloroacetic acid (CH2ClCOOH)
Substitute these values into the equilibrium expression and solve for [H+]. Since the concentration of chloroacetic acid is equal to the concentration of H+ and CH2ClCOO- ions, we can use x to represent this concentration:
Ka = x * x / (0.0286 - x)
Now, solve the equation for x.
With the value of x, you can now calculate the percent ionization:
Percent Ionization = (Concentration of dissociated ions / Initial concentration of acid) * 100
Substitute the concentration of dissociated ions (x) and the initial concentration of acid (0.0286 mol/1.60 L) into the formula to find the percent ionization.