Question

A beaker with 125 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 6.80 mL of a 0.300 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.

Answer 1

Ans. Step 1: Using HH equation:   pH = pKa + log ([A^-] / [AH])

            Where, AH = weak acid       ; A^- = Conjugate base

Putting the values in above equation-

            5.000 = 4.740 + log ([A^-] / [AH])

            Or, [A^-] / [AH] = antilog (5.000 – 4.740) = 10^-0.260

Or, [A^-] = 0.5495 [AH]       - equation 1

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

Comparing equation 1 and 2-

            [AH] + 0.5495 [AH] = 0.100

            Or, [AH] = 0.100 / 1.5495

            Hence, [AH] = 0.065

# Putting the value of [AH] in equation 2-

            [A^-] = 0.100 – 0.065 = 0.035

# Step 2:

A = Weak acid A-= Conjugate base Ka of acid1.80E-05 Kb 10-14/ Kb 5.56E-10 pKa -log Ka 4.74 Original Buffer Strong Acid [AH] = 0.065 M A0.035 M Vol.= 150.0 mL = [HCI Vol.- 0.300 M 6.8 mL0.007 L 0.150 L Using C1V1 (Original soln.) C2V2 (Mixed soln.) C1V1 Total vol. of mixed soln., V2 = 0.157 L Balanced Reaction H3OAH Initial [Conc] in mixed soln., C2- V2 H2O In mixed solution, Initial [AH]0.062 M In mixed solution, Initial [A10.033 M In mixed solution, Initial [HCI]0.013 M A1 consumed Initial [HCI]0.013 M Using HH equation IA] [AH 0.020 0.046 pH= pKa + log pOH-4.745 +log 4.389 Remaining [A (Initial Consumed)0.033 -0.013F0.020 [AH] formed [A] consumed0.013 Total [AH] (Initial Formed)0.033 0.013-0.046

# Change in pH = final pH – Initial pH = 4.389 – 5.000 = -0.611