(a) What is the pH of 50 mL of a 0.2M glycolic acid aqueous solution after adding 5 mL of 0.25 M KOH? The pKa of glycolic acis is 3.83 at 25 ∘C.

To find the pH of the solution, we need to understand the reactions that occur when glycolic acid (HA) reacts with KOH. First, let's start by writing the balanced chemical equation:

HA + KOH → H2O + KA

From the equation, we can see that one molecule of HA reacts with one molecule of KOH, producing water (H2O) and potassium glycolate (KA). Since the reaction is 1:1, we can determine the number of moles of glycolic acid and KOH present.

Given:
Volume of glycolic acid solution (V1) = 50 mL
Concentration of glycolic acid (C1) = 0.2 M
Volume of KOH solution (V2) = 5 mL
Concentration of KOH (C2) = 0.25 M

Now, let's calculate the moles of glycolic acid and KOH:

Moles of glycolic acid (moles_HA) = C1 * V1
= 0.2 M * 0.05 L (since 50 mL = 0.05 L)
= 0.01 moles

Moles of KOH (moles_KOH) = C2 * V2
= 0.25 M * 0.005 L (since 5 mL = 0.005 L)
= 0.00125 moles

Since the reaction between glycolic acid and KOH is 1:1, the moles of KOH consumed will be equal to the moles of HA reacted. Therefore, we have 0.00125 moles of glycolic acid remaining after the reaction.

Now, we can calculate the concentration of the remaining glycolic acid:

Concentration of glycolic acid after the reaction = Moles of glycolic acid remaining / Volume of solution after the reaction
= 0.00125 moles / (50 mL + 5 mL)
= 0.00125 moles / 0.055 L
≈ 0.0227 M

Next, we can use the Henderson-Hasselbalch equation to calculate the pH:

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

Given:
pKa = 3.83
[A-] (concentration of potassium glycolate) = Concentration of glycolic acid after the reaction = 0.0227 M
[HA] (concentration of glycolic acid) = 0.0227 M

pH = 3.83 + log(0.0227/0.0227)
= 3.83 + log(1)
= 3.83 + 0
= 3.83

Therefore, the pH of the solution after adding 5 mL of 0.25 M KOH to 50 mL of a 0.2 M glycolic acid solution is 3.83.