Final answer:
To reduce the pH of a buffer solution containing 0.0100 M acetic acid and 0.100 M sodium acetate to 4.75, approximately 1.02 mL of 10.0 M HNO3 must be added.
Step-by-step explanation:
A buffer solution is used to maintain a relatively constant pH when an acid or a base is added to a solution. To calculate how much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 4.75, we need to use the Henderson-Hasselbalch equation. The equation is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.
First, we need to find the pKa of acetic acid. The pKa is given by -log(Ka), where Ka is the acid dissociation constant. For acetic acid, Ka = 1.8x10⁻⁵, so pKa = -log(1.8x10^-5) = 4.74. Using the Henderson-Hasselbalch equation, we can calculate the desired concentration of the conjugate base:
pH = 4.75 = 4.74 + log([A-]/[HA])
log([A-]/[HA]) = 4.75 - 4.74 = 0.01
Now, we need to determine the ratio of [A-] to [HA]:
[A-]/[HA] = 10⁰·⁰¹ = 1.023
Since the buffer is made of acetic acid and sodium acetate, the ratio of their concentrations is equal to the ratio of [HA] to [A-]. Given that [HA] = 0.0100 M and [A-] = 0.100 M, we can set up the following equation:
0.0100 M / 0.100 M = 1.023
0.100 M = 1.023 * 0.0100 M
Now, we can solve for the desired concentration of the conjugate base:
[A-] = 1.023 * 0.0100 M
[A-] = 0.0102 M
Finally, we need to find out how much 10.0 M HNO3 must be added to achieve this concentration. Since the volume of the buffer is 1.00 L, we can use the equation:
C1V1 = C2V2
(10.0 M)(V1) = (0.0102 M)(1.00 L)
V1 = (0.0102 M)(1.00 L) / 10.0 M
V1 = 0.00102 L = 1.02 mL