how many mililitres of 0.800M KOH should be added to 5.02g of 1,2-pentanedoic acid (C5H8O4 FM 132.11) to give a pH 4.40 when diluted to 250ml

25.4mL of KOH 0.800M must be added to obtain the desire pH

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 =

can you list the answers so i can answer

the answer to this question is 3.