Question

A beaker with 2.00×102 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 6.90 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.

Answer 1

Answer: The pH will decrease by 0.109 units.

Step-by-step explanation:

To solve this problem, we will need to use the Henderson-Hasselbalch equation:

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

where pH is the initial pH of the buffer, pKa is the acid dissociation constant of acetic acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

First, we need to find the initial concentrations of [A-] and [HA] in the buffer. Since the total molarity of acid and conjugate base is 0.100 M and we know the volume of the buffer, we can use the following equation:

moles of acid = moles of conjugate base

0.100 M x 2.00x10^-2 L = [HA] x 2.00x10^-2 L

[HA] = 0.100 M

Since we know the pH of the buffer, we can use the following equation to find the concentration of the conjugate base:

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

5.000 = 4.740 + log([A-]/0.100)

[A-]/[HA] = 10^(5.000-4.740)

[A-]/[HA] = 1.995

[A-] = 1.995 x 0.100 M = 0.1995 M

Now, we need to find the new concentrations of [A-] and [HA] after the addition of HCl. Since the volume of the buffer is now 2.069x10^-2 L (2.00x10^-2 L + 6.90x10^-3 L), we can use the following equation:

moles of acid + moles of HCl = moles of conjugate base

0.100 M x 2.00x10^-2 L + 0.300 M x 6.90x10^-3 L = [HA] x 2.069x10^-2 L

[HA] = 0.1295 M

The concentration of the conjugate base can be found using the equation:

[A-]/[HA] = 10^(pH-pKa)

1.891 = 10^(pH-4.740)

pH-4.740 = log(1.891)

pH = log(1.891) + 4.740

pH = 5.000 - 0.109

Therefore, the pH will decrease by 0.109 units.