Monochloroacetic acid, HC2H2ClO2, is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of Ka for monochloroacetic acid is 1.35 10-3. Calculate the pH of a 0.22 M solution of monochloroacetic acid.

monochloroacetic acid is HA.

...............HA ==> H^+ + A^-
initial.......0.22....0......0
change.........-x.....x.......x
equil.......0.22-x....x.......x

Substitute into the expression for Ka and solve for H^+, then convert to pH.

To calculate the pH of a solution of monochloroacetic acid, we need to consider the acid dissociation reaction and the equilibrium constant (Ka) for the reaction.

The dissociation reaction of monochloroacetic acid, HC2H2ClO2, can be represented as follows:

HC2H2ClO2 (aq) ⇌ H+ (aq) + C2H2ClO2- (aq)

The equilibrium constant expression for the reaction is:

Ka = [H+] [C2H2ClO2-] / [HC2H2ClO2]

Given that the initial concentration of monochloroacetic acid is 0.22 M, we can assume that the change in concentration of H+ and C2H2ClO2- is negligible compared to the initial concentration. Therefore, we can assume that [H+] = x, [C2H2ClO2-] = x, and [HC2H2ClO2] = 0.22 - x.

Substituting these values into the equilibrium constant expression, we get:

Ka = x * x / (0.22 - x)

Given that Ka = 1.35 * 10^-3, we can set up the following equation:

1.35 * 10^-3 = x * x / (0.22 - x)

Solving this equation will give us the value of x, which represents the concentration of H+ in the solution. Let me solve this equation step-by-step to calculate the pH.

To calculate the pH of a solution of monochloroacetic acid, we need to use the dissociation constant, Ka. The dissociation constant, Ka, for an acid is a measure of its strength and represents its tendency to donate a proton (H+ ion) in water.

The equation for the dissociation of monochloroacetic acid, HC2H2ClO2, is:
HC2H2ClO2 (aq) ⇌ H+ (aq) + C2H2ClO2- (aq)

The equilibrium expression for this dissociation is:
Ka = [H+] [C2H2ClO2-] / [HC2H2ClO2]

Given that the initial concentration of monochloroacetic acid is 0.22 M, we can assume that x is the concentration of H+ and C2H2ClO2- at equilibrium, and 0.22 - x is the concentration of HC2H2ClO2 at equilibrium.

Using the expression for Ka and substituting the known values, we have:
1.35 x 10^-3 = x * (0.22 - x) / (0.22 - x)

Since x is assumed to be small compared to 0.22, we can approximate (0.22 - x) as 0.22. The equation becomes:
1.35 x 10^-3 = x * 0.22 / 0.22

Simplifying the equation further, we get:
1.35 x 10^-3 = x

Now, we can solve for x by taking the square root of both sides of the equation:
x = √(1.35 x 10^-3)

x ≈ 0.0368

The concentration of H+ ions in the solution is approximately 0.0368 M.

To calculate the pH of the solution, we can take the negative logarithm (base 10) of the H+ concentration:
pH = -log[H+]

pH ≈ -log(0.0368)

pH ≈ 1.434

Therefore, the pH of a 0.22 M solution of monochloroacetic acid is approximately 1.434.