Respuesta :
Solution:
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.
Solution:
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.
