Question

Consider a 0.60-M solution of HC₃H₅O₃, lactic acid (K_a = 1.4 × 10^-4).

a. Which of the following are major species in the solution?

i. HC₃H₅O₃

ii. C₃H₅O₃^-

iii. H⁺

iv. H₂O

v. OH^-

b. Complete the following ICE table in terms of x, the amount (mol/L) of lactic acid that dissociates to reach equilibrium.

	HC3H5O3	[H+]	C3H5O3-
Initial
Change
Equilibrium	0.60-x

c. What is the equilibrium concentration for${\mathbf{C}}_3{\mathbf{H}}_5{{\mathbf{O}}_3}^{-}$?
d. Calculate the pH of the solution.

Short answer

a. The major species present in the solution are i, ii, and iii.

b.

	HC3H5O3	[H+]	C3H5O3-
Initial	0.6	0	0
Change	-x	+x	+x
Equilibrium	0.60-x	x	x

c.$\left[{\mathrm{C}}_3{\mathrm{H}}_5{{\mathrm{O}}_3}^{-}\right]=9.1\times {10}^{-3}$

d. pH=2.04

Step-by-step solution

Major species present in solution

Lactic acid is a weak acid but much stronger than water, and it dissociates as:

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\rightleftarrows {\text{\hspace{0.17em}H}}^{\text{+}}{\text{+\hspace{0.17em}C}}_{\text{3}}{\text{H}}_{\text{5}}{{\text{O}}_{\text{3}}}^{\text{-}}$

So the major species present in the solution are i, ii, and iii.

Lactic Acid Dissociation

HC₃H₅O₃dissociates in water as:

$\begin{array}{l}{\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\rightleftharpoons {\text{\hspace{0.17em}H}}^{\text{+}}{\text{+\hspace{0.17em}C}}_{\text{3}}{\text{H}}_{\text{5}}{{\text{O}}_{\text{3}}}^{\text{-}}\\ \mathrm{Initial}0.600\\ \text{Change\hspace{1em}\hspace{1em}}-\mathrm{x}+\mathrm{x}+\mathrm{x}\\ \text{Equilibrium\hspace{0.17em}}0.6-\mathrm{x}\mathrm{x}\mathrm{x}\end{array}$

	HC3H5O3	[H+]	C3H5O3-
Initial	0.6	0	0
Change	-x	+x	+x
Equilibrium	0.60-x	x	x

Equilibrium Concentration of C3H5O3-

From Ka, the equilibrium concentration of C₃H₅O₃^-can be calculated as:

$\begin{array}{c}{\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{{\text{O}}_{\text{3}}}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\mathrm{H}{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]}\\ 1.4\times {10}^{-4}=\frac{{\mathrm{x}}^2}{0.6-\mathrm{x}}\\ \mathrm{x}=9.1\times {10}^{-3}\end{array}$

[C₃H₅O₃^-] = $\mathrm{x}=9.1\times {10}^{-3}$

pH Calculation

pH from H⁺ ion concentration:

${\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]=-log(}9.1\times {10}^{-3}\text{)=2.04}$