Respuesta :
First, we have to use Henderson-Hasselblach equation to get the weak base/weak acid ratio:
PH = Pka + ㏒[A]/[HA]
when PH = 4
Pka = 4.2
A is the benzoate
HA is the benzoic acid
so,
4 = 4.2 + ㏒[A]/[HA]
∴[A]/[HA] = 0.631
when [A]/[HA] = 0.12 m * VA / 0.1 m * VHA
∴0.631 = 0.12m* VA / 0.1m * VHA
now, when we have the total volume of the solution is 100 mL
∴VA = 100 mL - VHA ( we can substitute with 100 - VHA instead of VA)
∴0.631 = 0.12 * (100-VHA) / 0.1 * VHA
we can assume that VHA = X
so,
0.631 = 0.12 * (100-X) / 0.1 * X by solving this equation for X
∴X = 65.5
∴VHA = 65.5 mL
so, VA = 100mL - 65.5 ML
= 34.5 mL
PH = Pka + ㏒[A]/[HA]
when PH = 4
Pka = 4.2
A is the benzoate
HA is the benzoic acid
so,
4 = 4.2 + ㏒[A]/[HA]
∴[A]/[HA] = 0.631
when [A]/[HA] = 0.12 m * VA / 0.1 m * VHA
∴0.631 = 0.12m* VA / 0.1m * VHA
now, when we have the total volume of the solution is 100 mL
∴VA = 100 mL - VHA ( we can substitute with 100 - VHA instead of VA)
∴0.631 = 0.12 * (100-VHA) / 0.1 * VHA
we can assume that VHA = X
so,
0.631 = 0.12 * (100-X) / 0.1 * X by solving this equation for X
∴X = 65.5
∴VHA = 65.5 mL
so, VA = 100mL - 65.5 ML
= 34.5 mL
The volume of 0.100 M benzoic acid required is 65.6 mL and the volume of 0.120 M sodium benzoate is 34.4 mL.
Buffer solutions
Buffer solutions are solutions that resists changes in its pH when a little quantity of acid or base is added to it them.
The Henderson-Hasselblach equation
The Henderson-Hasselblach equation is used to determine the weak base/weak acid ratio in a buffer solution
- Henderson-Hasselblach equation is given as: pH = Pka + ㏒[A]/[HA]
From the data provided:
pH of buffer = 4.00
pKa of acid = 4.20
- A is benzoate and benzoic acid is HA
Using the Henderson-Hasselbach equation:
4.00 = 4.20 + ㏒[A]/[HA]
[A]/[HA] = 0.63
- Also; Molarity = number of moles × volume
[A] = 0.12 × V1
[HA] = 0.10 × V2
Hence:
0.1 2 × V1 / 0.10 × V2 = 0.63
V1 = 0.10 × 0.63 × V2 / 0.12
V1 = 0.525 × V2------- eqn 1
Also, since V1 + V2 = 100 mL
V2 = 100 - V1
Substitute V2 = 100 - V1 in eqn 1
V1 = 0.525 × {100 - V1}
V1 = 52.5 - 0.525V1
1.525V1 = 75.6
V1 = 52.5 ÷ 1.525
V1 = 34.4 mL
Then V2 = 100 - 34.4
V2 = 65.6
Therefore, the volume of 0.100 M benzoic acid required is 65.6 mL and the volume of 0.120 M sodium benzoate is 34.4 mL.
Learn more about buffers and Henderson-Hasselblach equation at: https://brainly.com/question/11851669