Answer to Question #151295 in Chemistry for W

Question

A 0.0560g sample of acetic acid is dissolved in enough water to make 50.0mL of solution. Calculate the equilibrium concentrations of each of the species in the reaction and determine the pH of the solution.

Answer 1

When the acetic acid is dissolved in water, the following dissociation reaction takes place:

HAc $\leftrightarrow$ H⁺_aq + Ac^-_aq.

The total concentration of the acetic acid species (HAc and Ac^-) is (the number of the moles of HAc is its mass divided by its molar mass M=60.05 g/mol):

$c = \frac{n}{V} = \frac{m}{MV} = \frac{0.0560}{60.05\cdot50\cdot10^{-3}} = 0.01865$ mol/L.

Therefore, after the dissociation reaction comes to the equilibrium, if the equilibrium concentration of the acetate anion is $x$ , the concentration of protons is also $x$ and the concentration of HAc is $c-x$ .

The equilibrium constant of the dissociation reaction of the acetic acid is 1.8x10^-5:

$K_{eq} = \frac{[H^+][Ac^-]}{[HAc]} = \frac{x\cdot x}{c-x} = 1.8\cdot10^{-5}$ .

Solving this equation gives the value of $x$ :

$x^2+1.8\cdot10^{-5}x - 1.8\cdot10^{-5}c = 0$

$x = 5.7\cdot10^{-4}$

Therefore:

$[H^+]=[Ac^-] = 5.7\cdot10^{-4}$ M and

$[HAc] = 0.01865 - 5.7\cdot10^{-4} = 1.81\cdot10^{-2}$ M.

Finally, the pH of the solution is:

$pH = -\text{log}[H^+] = -\text{log}(5.7\cdot10^{-4}) = 3.24$

Answer: the equilibrium concentrations of each of the species in the reaction are 5.7x10^-4M for H⁺_aq and Ac^-_aq and 1.81x10^-2M for HAc. The pH of the solution is 3.24.