Answer:
The degree of ionization = 8.56%; pH = 1.4264; pOH = 12.5736.
Explanation:
The balanced equilibrium reaction for the dissociation of lactic acid is given below;
CH3CH(OH)COOH ( aq ) + H2O <----------> CH3CH(OH)COO^-(aq) + H3O^+.
(It must be noted that CH3CH(OH)COOH is a weak and monoprotic acid).
==> The initial concentration of CH3CH(OH)COOH = 0.1051 M and the initial concentration of H3O^+ and CH3CH(OH)COO^- = 0 M respectively.
==> At equilibrium, the concentration of CH3CH(OH)COOH = 0.1051 - x M and the concentration of CH3CH(OH)COOH is x M and the concentration of H3O^+ = x M.
Hence, the formula relating the ka value and the concentration of the species in the Reaction is given below;
Ka = [H3O^+] [CH3CH(OH)COO^-] / [CH3CH(OH)COOH].
8.40 × 10^-4 = (x^2)/ 0.1051 - x.
x^2 = (8.40 × 10^-4 × 0.1051) - 8.40 × 10^-4 x.
x^2 + 8.40 × 10^-4 - 0.000088284.
Therefore, solving the quadratic equation for the equation above, the value of x= 0.0089851262617787( we are ignoring the x value that has negative sign attached to it)
x= 0.009 M
(1). The degree of ionization= [concentration of ionized molecules/ Initial concentration of the acid] × 100.
The degree of ionization = (0.009/ 0.1051) × 100.
= 8.56%.
(2). In order to calculate the value of pH we make use of Henderson-Hasselbalch Equation which is given below;
pH = pKa + log ( [ CH3CH(OH)COO^-] / [CH3CH(OH)COOH] ).
pH= 3.1 + log ( 0.009/ 0.1051).
pH= 3.1 + log 0.08563273073.
pH = 3.1 + (-1.0673602066).
pH= 3.1 - 1.6736.
pH = 1.4264.
(Recall that pKa = - log ka).
(3). pH + pOH = 14.
pOH = 14 - pH.
pOH = 12.5736.
