Chloroacetic acid has a relatively large equilibrium constant, so at low acid concentrations it is necessary to use the quadratic equation in order to calculate the concentrations of the aqueous species. Ka for ClCH2COOH is 1.4 ´ 10-3. For the problems that follow, consider one liter of a solution that originally had 0.0200 mol of chloroacetic acid.
What is the equilibrium concentration of [ClCH2COOH]?
a. 0.010
b. 0.012
c. 0.015
d. 0.017
ClCH2COOH ==> ClCH2COO^- + H^+
Ka = (H^+)(ClCH2COO^-)/((ClCH2COOH)
(H^+)=x
(ClCH2COO^-)=x
(ClCH2COOH) = 0.02-x
a. 0.010
The answer is C).015
Find x through the quadratic equation x^2^ +(1.4E-3)x - (2.8E-5)=0
To find the equilibrium concentration of [ClCH2COOH], we can use the quadratic equation, since the equilibrium constant for chloroacetic acid (Ka) is relatively large.
Step 1: Write the balanced equation for the dissociation of chloroacetic acid:
ClCH2COOH ⇌ ClCH2COO- + H+
Step 2: Set up an ICE table (Initial, Change, Equilibrium) to track the concentrations of the species involved in the equilibrium.
Initial:
[ClCH2COOH] = 0.0200 mol/L
[ClCH2COO-] = 0 mol/L (since it starts undissociated)
[H+] = 0 mol/L (since it starts undissociated)
Change:
Let's assume that x mol/L of chloroacetic acid dissociates.
Since chloroacetic acid has a 1:1 stoichiometry for the dissociation, [ClCH2COO-] and [H+] will both increase by x mol/L.
Equilibrium:
[ClCH2COOH] = 0.0200 - x mol/L
[ClCH2COO-] = x mol/L
[H+] = x mol/L
Step 3: Write the equilibrium expression for the dissociation of chloroacetic acid using the concentrations from the equilibrium row of the ICE table:
Ka = ([ClCH2COO-] * [H+]) / [ClCH2COOH] = (x * x) / (0.0200 - x)
Step 4: Substitute the value of Ka given in the problem (1.4´10-3) into the equilibrium expression:
1.4´10-3 = (x^2) / (0.0200 - x)
Step 5: Rearrange the equation to solve for x:
1.4´10-3 * (0.0200 - x) = x^2
Step 6: Simplify the equation:
0.028 - 1.4´10-3x = x^2
Step 7: Rearrange the equation to have a quadratic equation format:
x^2 + 1.4´10-3x - 0.028 = 0
Step 8: Solve the quadratic equation using the quadratic formula or factoring. In this case, since the quadratic equation can't be easily factored, we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 1.4´10-3, and c = -0.028.
x = (-1.4´10-3 ± √((1.4´10-3)^2 - 4(1)(-0.028))) / (2(1))
x = (-1.4´10-3 ± √(1.96´10-6 + 0.112)) / 2
Simplifying further, we get:
x ≈ (-1.4´10-3 ± √0.113) / 2
Step 9: Calculate the two possible values of x and choose the value that is physically meaningful. In this case, since we're looking for the equilibrium concentration of [ClCH2COOH], x must be less than 0.0200 mol/L.
x ≈ (-1.4´10-3 - √0.113) / 2 ≈ -0.0074 (not valid)
x ≈ (-1.4´10-3 + √0.113) / 2 ≈ 0.0147
Since the negative value is not valid, we choose the positive value of x, approximately 0.0147 mol/L, which represents the equilibrium concentration of [ClCH2COOH].
Therefore, the equilibrium concentration of [ClCH2COOH] is approximately 0.0147 mol/L.
Among the given answer choices (a, b, c, d), the closest value to 0.0147 mol/L is 0.015 mol/L (c).