Question

Can the italicized portion of the problem be explained? I understand up until the 0.015 moles of protonated acetic acid remained and I understand the HH portion of the problem after but I would like some clarity on how this number was concluded.

2. The following question has two parts.
a) What is the final pH of a solution obtained by mixing 250 ml of 0.3 M acetic acid with 300 ml

of 0.2 M KOH? (pKb of acetate = 9.24). (Show your work!)

Moles of acetic acid = 0.25 l X 0.3 M = 0.075 moles
Adding 0.3 l X 0.2 M = 0.06 moles of OH- to this solution will convert 0.06 moles of acetic

acid to 0.06 moles of acetate, 0.015 moles of protonated acetic acid will remain. pKa = 14 - pKb = 14 - 9.24 = 4.76

pH = pKa + log [CH3CO2-][CH3CO2H]

= 4.76 + log 0.06 moles / 0.55 l 0.015 moles / 0.55 l

= 4.76 + log 0.06 moles0.015 moles

= 5.36

pH = 5.36

Answer 1

Ans. Clarification 1: In the final mixed solution, you have –

            (0.06 mol / 0.55 L) = 0.1091 M of acetate ion

            (0.015 mol / 0.55 L) = 0.0273 M of acetic acid

# when you divide (0.06 mol / 0.55 L) by (0.015 mol / 0.55 L) , you get the same ratio as (0.1091 / 0.0273); and the same would be the result, too. However, it’s desirable to calculate the concertation in terms of molarity because the HH equation is based on it.

# The calculation of concentration in terms of MOLARITY is crucial because HH equation uses the amount of weak acid and its conjugate base in terms of molarity.

# Clarification 2: Acetic acid is weak acid. Ka is the acid dissociation constant, and pKa is –ve log of Ka (note that pKa is used in HH equation).

SO, in order to calculate pH, we first need to calculate pKa of the weak acid.

            pKa + pKb = 14.00

            Or, pKa = 14.00 – pKb = 14.00 – 9.24 = 4.76

Now, calculate pH of solution using HH equation as follow-

            Where, AH = weak acid (acetic acid)        

; A- = conjugate base (acetate ion)

# Clarification 3: In the final mixed solution, you have –

            (0.06 mol / 0.55 L) = 0.1091 M of acetate ion

            (0.015 mol / 0.55 L) = 0.0273 M of acetic acid

Where, acetate ion (A^-) can be treated as BASE and acetic acid can be treated as its conjugate acid. So, you can also use HH equation for BASE to calculate the pH of resultant solution using the pKb value as follow-