Question

What are the concentrations of acetic acid (pka = 4.76) and acetate in a buffer solution of 0.10 m at ph 4.9?

Answer 1

We have to use the Henderson-Hasselbalch equation: for this calculation

Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity by using pKa, the negative log of the acid dissociation constant in biological and chemical systems. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid-base reaction.

The equation is given by:

Here, [HA] is the molar concentration of the un dissociated weak acid, [A⁻] is the molar concentration (molarity, M) of this acid's conjugate base and pKa is −log10 Ka where Ka is the acid dissociation constant, that is:

pH = pKa + log([A^-]/[HA])

We look up the pKa for acetic acid:

pKa = 4.76

Let x = molarity of AcO^- and y = molarity of AcOH: Then we have the following two equations in two unknowns:

(1) x + y = 0.10 M

and

(2) 4.9 = 4.76 + log(x/y)

Further calcite the value of x and y by algebraic method and get the answer.

Answer 2

We have to use the Henderson-Hasselbalch equation: for this calculation

Henderson–Hasselbalch equation describes the derivation of pH as a measure of acidity by using pKa, the negative log of the acid dissociation constant in biological and chemical systems. The equation is also useful for estimating the pH of a buffer solution and finding the equilibrium pH in acid-base reaction.

The equation is given by:

Here, [HA] is the molar concentration of the un dissociated weak acid, [A⁻] is the molar concentration (molarity, M) of this acid's conjugate base and pKa is −log10 Ka where Ka is the acid dissociation constant, that is:

pH = pKa + log([A^-]/[HA])

We look up the pKa for acetic acid:

pKa = 4.76

Let x = molarity of AcO^- and y = molarity of AcOH: Then we have the following two equations in two unknowns:

(1) x + y = 0.10 M

and

(2) 4.9 = 4.76 + log(x/y)

Further calcite the value of x and y by algebraic method and get the answer.