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 5.05
help please :(
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to its acid/base components.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
- pH is the desired pH (5.05 in this case)
- pKa is the acid dissociation constant of the weak acid (acetic acid in this case)
- [A-] is the concentration of the conjugate base (sodium acetate)
- [HA] is the concentration of the weak acid (acetic acid)
First, we need to calculate the pKa of acetic acid. The pKa value is often provided in a chemical reference book or can be found online. For acetic acid, the pKa value is approximately 4.76.
Next, we substitute the known values into the Henderson-Hasselbalch equation and solve for the log ([A-]/[HA]) term:
5.05 = 4.76 + log ([A-]/[HA])
Now, let's rearrange the equation and solve for ([A-]/[HA]):
log ([A-]/[HA]) = 5.05 - 4.76
= 0.29
Next, we calculate the ratio of [A-]/[HA]. Since we are using sodium acetate as the conjugate base, and acetic acid as the weak acid, the ratio will be:
[A-]/[HA] = [sodium acetate]/[acetic acid]
= 0.100 M / 0.0100 M
= 10
Now, we substitute this value back into the equation and solve for the concentration of acetic acid ([HA]) required:
log (10) = 0.29
Using logarithmic properties, we find that 10 is equal to 10^0.29. Therefore:
10^0.29 = ([A-]/[HA])
1.906 = ([A-]/[HA])
Since the concentration of sodium acetate ([A-]) is 0.100 M, we can calculate the concentration of acetic acid needed:
0.100 M / 1.906 = [HA]
0.0525 M = [HA]
So, you will need to add 0.0525 M of 10.0 M HNO3 to 1.00 L of the buffer to achieve a pH of 5.05.