chem find the mol for a solution

Question

To a 400 mL solution of 0.5 M acetic acid (Ka = 1.75× 10⁻⁵), you add 0.1 moles of KOH to prepare a buffer solution. If you then mix in 100 mL of 0.3 M HCl, what will the final pH be? Hint: the final pH would be the same if you first added the KOH to the HCl and then combined the 400 mL and 100 mL solutions.

Answer 1

Acetic acid = CH3COOH

moles CH3COOH = 400 ml x 1 L / 1000 ml x 0.5 mol / L = 0.20 mols

moles KOH = 0.1 mols

CH3COOH + KOH <==> CH3COOK + H2O

0.2.................0.1...................0.................Initial

-0.1...............-0.1.................+0.1..............Change

0.1...................0.....................0.1.............Equilibrium

If you now add 100 ml of 0.3 M HCl, we are adding 0.1 L x 0.3 mol / L = 0.03 mol HCl (H⁺). The H⁺ will react with the conjugate base, CH3COO- to produce CH3COOH:

CH3COO- + H+ ==> CH3COOH

0.1..............0.03.............0.1...............Initial

-0.03...........-0.03...........+0.03...........Change

0.07................0.................0.13.........Equilibrium

Final volume = 400 ml + 100 ml = 500 ml = 0.5 L

Final [CH3COO-] = 0.07 mol / 0.5 L = 0.14 M

Final [CH3COOH] = 0.13 mol / 0.5 L = 0.26 M

Henderson Hasselbalch equation:

pH = pKa + log [base] / [acid]

pH = ?

pKa = -log Ka = -log 1.75x10^-5 = 4.76

pH = 4.76 + log (0.14 / 0.26)

pH = 4.76 -0.27

pH = 4.49

If you do it by combining KOH and HCl first, as suggested in the hint, then you have the following:

KOH + HCl == KCl + H2O

moles KOH = 0.1 mols

moles HCl = 0.03 mols

Excess mols KOH = 0.07 mols KOH

CH3COOH + KOH ==> CH3COOK + H2O

0.2..................0.07..................0..............Initial

-0.07..............0.07................+0.07..........Change

0.13.................0....................0.07...........Equilibrium

pH = pKa + log [base] / [acid]

pH = 4.76 + log (0.07 / 0.13)

pH = 4.76 - 0.27

pH = 4.49