What is the percent ionization of a solution prepared by dissolving 0.0286mol of chloroacetic acid, in 1.60L of water?for choloroacetic acid, Ka=1.4x10^-3

Question

What is the percent ionization of a solution prepared by dissolving 0.0286mol of chloroacetic acid, in 1.60L of water?for choloroacetic acid, Ka=1.4x10^-3

Answer 1

http://www.jiskha.com/display.cgi?id=1363711504

Answer 2

To calculate the percent ionization of a solution, we need to first determine the concentration of the ionized species. In this case, chloroacetic acid (CH2ClCOOH) will ionize into its conjugate base, chloroacetate ion (CH2ClCOO-), and a hydrogen ion (H+). The equilibrium equation for this reaction can be written as follows:

CH2ClCOOH ⇌ CH2ClCOO- + H+

The Ka value, which is the acid dissociation constant, represents the extent of the ionization. In this case, the Ka for chloroacetic acid is given as 1.4x10^-3.

To calculate the percent ionization, we need to find the equilibrium concentration of CH2ClCOO- and H+ ions. Since the initial concentration of chloroacetic acid (CH2ClCOOH) is 0.0286 mol and the solution volume is 1.60 L, the initial concentration of chloroacetic acid can be calculated as:

Initial concentration of CH2ClCOOH = 0.0286 mol / 1.60 L = 0.0179 mol/L

Let x be the concentration of CH2ClCOO- and H+ ions formed at equilibrium.

Using an ICE table (Initial, Change, Equilibrium), we can set up the following equations:

CH2ClCOOH ⇌ CH2ClCOO- + H+
Initial: 0.0179 0 0
Change: -x +x +x
Equilibrium: 0.0179 - x x x

Since the change in concentration is likely to be very small compared to the initial concentration, we can assume that 0.0179 - x ≈ 0.0179.

Now, we can express the equilibrium constant (Ka) using the concentrations of the species involved:

Ka = [CH2ClCOO-][H+] / [CH2ClCOOH]

Substituting the equilibrium concentrations into the expression, we get:

1.4x10^-3 = x^2 / (0.0179 - x)

Note that we neglected the concentration of x in the denominator because it is very small compared to 0.0179.

Simplifying the equation and solving for x, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Applying this formula with a = 1, b = -1.4x10^-3, and c = -1.4x10^-3, we find:

x = 0.0388 mol/L

Now, we can calculate the percent ionization by dividing the concentration of ionized species (x) by the initial concentration of the acid and multiplying by 100:

Percent Ionization = (0.0388 mol/L / 0.0179 mol/L) x 100 = 216.2%

Therefore, the percent ionization of the solution is 216.2%.