Question

In an aqueous solution of an acid (or a base), there is an equilibrium between the acid HA and the products of its dissociation — H+ and A−: HA H+ + A−. This equilibrium can be described using a dissociation constant: Ka = c(H+)·c(A−) c(HA) , where c is molar concentration. The values of Ka for various acids are known. They show how far the dissociation goes and how strong the acid is. For strong acids, it can be of the order of 109 (molar concentrations are assumed in mol/L, and the units of Ka are dropped). In calculations, it is often convenient to use the degree of dissociation, α, which is a fraction of the acid molecules that have dissociated. For acetic acid, Ka = 1.8·10−5. Evaluate the concentration of hydrogen cations in its 0.010M aqueous solution.
Do it rigorously using the degree of dissociation and then approximately assuming that in equilibrium, the concentration of 2 of 3 intact acetic acid molecules is very close to 0.010M.
Comment on your findings.

Answer 1

Answer:

(1) 4.2 × 10⁻⁴ mol·L⁻¹; (2) 4.3 × 10⁻⁴ mol·L⁻¹; (3)  The results agree.

Explanation:

1. Exact solution using α

                     HA + H₂O ⇌ H₃O⁺     +     A⁻

I/mol·L⁻¹:     0.010                0                 0

C/mol·L⁻¹:   -0.010α        +0.010α      +0.010α

E/mol·L⁻¹:  0.010(1-α)        0.010α        0.010α

$K_{\text{a}} = \dfrac{\text{[H _{3} O ^{+} ][A ^{-} ]}}{\text{[HA]}} = 1.8 \times 10^{-5}$

$\begin{array}{rcl}\dfrac{\text{[H _{3} O ^{+} ][A ^{-} ]}}{\text{[HA]}}& = &1.8 \times 10^{-5}\\\\ \dfrac{0.010\alpha\times 0.010\alpha }{0.010(1-\alpha)}& = &1.8 \times 10^{-5}\\\\ 0.000100\alpha^{2} & = & 0.010(1-\alpha)\times1.8 \times 10^{-5} \\& = &(0.010-0.010\alpha) \times 1.8 \times 10^{-5}\\& = & 1.8 \times 10^{-7} - 1.8 \times 10^{-7}\alpha\\0.000100\alpha^{2} + 1.8 \times 10^{-7}\alpha - 1.8 \times 10^{-7}& = & 0\\\alpha^{2}+ 0.00180\alpha - 0.00180& = & 0\\\end{array}$

a = 1; b = 0.00180; c = -0.00180

Solve using the quadratic formula

α = 0.041 536

[H₃O⁺] = 0.010× 0.0415 mol·L⁻¹ = 4.2 × 10⁻⁴ mol·L⁻¹

2. Approximate solution assuming x is negligible

                    HA + H₂O ⇌ H₃O⁺     +     A⁻

 I/mol·L⁻¹:  0.010                  0               0

C/mol·L⁻¹:     -x                    +x             +x

E/mol·L⁻¹:   0.010                  x               x

$\begin{array}{rcl}\dfrac{\text{[H _{3} O ^{+} ][A ^{-} ]}}{\text{[HA]}}& = &1.8 \times 10^{-5}\\\\ \dfrac{x\times x}{0.010}& = &1.8 \times 10^{-5}\\\\ x^{2} & = & 0.010\times1.8 \times 10^{-5} \\& = &1.8 \times 10^{-7}\\x & = & \sqrt{1.8 \times 10^{-7}}\\& = & \mathbf{4.3 \times 10^{-4}}\\\end{array}$

[H₃O⁺] = 0.010× 0.0415 mol·L⁻¹ = 4.3 × 10⁻⁴ mol·L⁻¹

3. Compare the results

The initial concentrations were known to two significant figures. The results agree to two significant figures ( ±1 unit in the second significant figure).