Calculate the ratio of NaF to HF required to create a buffer with pH=3.95.
Also, can you please explain in a nutshell the whole concept of buffers?
Thanks!
Never mind, I figured it out.
what was the anwer?
thanks!!
wait what? that's not right the -log(.494) = 2.02
To calculate the ratio of NaF to HF required to create a buffer with pH=3.95, you need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation for a buffer is:
pH = pKa + log([A-]/[HA])
In this equation, pH is the desired pH of the buffer, pKa is the acid dissociation constant of the weak acid (HF in this case), [A-] is the concentration of the conjugate base (NaF in this case), and [HA] is the concentration of the weak acid (HF).
To solve for the ratio of [A-]/[HA], rearrange the Henderson-Hasselbalch equation:
pH - pKa = log([A-]/[HA])
10^(pH - pKa) = [A-]/[HA]
Now, substitute the values and solve for [A-]/[HA]:
10^(3.95 - 3.2) = [A-]/[HA]
1.778 = [A-]/[HA]
This means that the ratio of NaF to HF required to create a buffer with pH=3.95 is approximately 1.778.
Now, moving on to the concept of buffers:
Buffers are solutions that resist changes in pH when small amounts of acids or bases are added to them. They are made up of a weak acid and its conjugate base (or a weak base and its conjugate acid). The key principle behind buffers is the ability of the weak acid or weak base to react with any added H+ or OH- ions, which helps to maintain the overall pH of the solution.
When a small amount of acid is added to a buffer solution, the weak acid in the buffer reacts with the added H+ ions to prevent a significant decrease in pH. Similarly, when a small amount of base is added, the conjugate base of the weak acid in the buffer reacts with the added OH- ions to prevent a significant increase in pH.
Buffers are widely used in various applications, such as biological systems (e.g., maintaining the pH of blood), chemical reactions (e.g., controlling the pH in enzymatic reactions), and analytical chemistry (e.g., calibration of pH meters). They are essential in maintaining the stability and functionality of many processes in both nature and the laboratory.
john the answer is right 3.1; u have to key on ur calc 2nd key then log key then the (^)key :) to get that answer.
pH=pKa + log(base/acid)
3.95=3.465 + log(base/acid)
.494=log(base/acid)
10^.494
3.1