A beaker with 160102 mL of an acetic acid buffer with a pH o

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.60 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( ) sign if the pH has decreased.

Solution

Ans. Step 1: Calculate Moles of acetic acid and acetate ion in original buffer:

Using Henderson- Hasselbalch equation -

            pH = pKa + log ([A^-] / [AH])                    - equation 1

                        where, [A^-] = acetate ions = [CH₃COO^-]

[AH] = Acetic acid

pKa of acetic acid = 4.74

Putting the values in equation 1-

            5.00 = 4.75 + log ([A^-] / [AH])

Or, 5.00 – 4.75 = log ([A^-] / [AH])

            Or, ([A^-] / [AH]) = 10^0.25 = 1.778

            Or, [A^-] = 1.778 [AH]                      - equation 1

Given, [A^-] + [AH] = 0.100 M                   - equation 2

# Putting the value of [A^-] in equation 2

            1.778 [AH] + [AH] = 0.100 M

            Or, [AH] = 0.100 M / 2.778

            Hence, [AH] = 0.036 M

Putting the value of [AH] in equation 2-

            [A^-] = 0.100 M – [AH] = 0.100 M – 0.036 M = 0.064 M

Therefore, [AH] = [CH3COOH] = 0.036 M

                        [A^-] = [CH3COO^-] = 0.064 M

# Now,

            Moles of CH₃COOH = Molarity x Volume of solution in liters

                                                = 0.036 M x 0.160 L

                                                = 0.00576 mol

            Moles of CH₃COO^- = 0.064 M x 0.160 M = 0.01024 mol

# Step 2: Accounting the change with addition of NaOH:

# Moles of NaOH added = 0.410 M x 0.00460 L = 0.001886 mol

# Balanced reaction: CH₃COOH + NaOH ------> Na⁺ + CH₃COO^- + H₂O

Stoichiometry: 1 mol NaOH neutralizes 1 mol CH₃COOH and forms 1 mol CH₃COO^-

So, moles of CH₃COOH neutralized = Moles of CH₃COO^- formed = 0.001886 mol

# Now,

            Remaining moles of CH₃COOH = Moles in original buffer – Moles neutralized

                                                            = 0.00576 mol - 0.001886 mol

                                                            = 0.003874 mol

            Remaining moles of CH₃COO^- = Moles of original buffer + Moles formed

                                                            = 0.01024 mol + 0.001886 mol

                                                            = 0.012126 mol

# Step 3: Calculating pH after Change:

So far, we have-

            Remaining moles of CH₃COOH = 0.003874 mol

            Remaining moles of CH₃COO^- = 0.012126 mol

Total volume of solution after mixing = 160.0 mL + 4.60 mL

= 164.60 mL = 0.1646 L

Now, in the final solution-

            [CH₃COOH] = 0.003874 mol / 0.1646 L = 0.02354 M

            [CH₃COO^-] = 0.012126 mol / 0.1646 L = 0.07367 M

# Again, putting the final concentrations in HH equation-

            pH = 4.74 + log (0.07367 / 0.02354) = 4.74 + log 3.1296 = 4.74 + 0.49 = 5.23

Therefore, new pH after addition of NaOH = 5.23