Answer:
25.4mL of KOH 0.800M must be added to obtain the desire pH
Step-by-step explanation:
pKa of 1,2-pentanedioic acid is 4.34.
It is possible to find pH of a buffer by using H-H equation, thus:
pH = pka + log [A⁻] / [HA]
Where [HA] is concentration of acid and [A⁻] is concentration of conjugate acid.
Replacing:
4.40 = 4.34 + log [A⁻] / [HA]
1.148 = [A⁻] / [HA] (1)
5.02g of 1,2-pentanedioic acid (Molar mass: 132.11g/mol) are:
5.02g ₓ (1mol / 132.11g) = 0.0380 moles of acid. That means:
[A⁻] + [HA] = 0.0380 (2)
Replacing (1) in (2):
1.148 = 0.0380 - [HA] / [HA]
1.148[HA] = 0.0380 - [HA]
2.148[HA] = 0.0380
[HA] = 0.0177 moles
Thus:
[A⁻] = 0.0380 - 0.0177 = 0.0203 moles [A⁻]
The moles of A⁻ comes from the reaction of the weak acid with KOH, that is:
HA + KOH → A⁻ + H₂O + K⁺
Thus, you need to add 0.0203 moles of KOH to produce 0.0203 moles of A⁻. As KOH solution is 0.800M:
0.0203 moles KOH ₓ (1L / 0.800mol) = 0.0254L of KOH 0.800M =
25.4mL of KOH 0.800M must be added to obtain the desire pH