Of the following solutions, which has the greatest buffering capacity?

A) 0.821 M HF and 0.909 M NaF
B) They are all buffer solutions and would all have the same capacity.
C) 0.100 M HF and 0.217 M NaF
D) 0.121 M HF and 0.667 M NaF
E) 0.821 M HF and 0.217 M NaF

how do i know when something has greater buffer capacity?

p.s. happy thanksgiving to everyone! sucks to be studying today of all days!

To determine which solution has the greatest buffering capacity, we need to consider the Henderson-Hasselbalch equation, which is given by:

pH = pKa + log([salt]/[acid])

In this equation, pKa represents the negative logarithm of the acid dissociation constant, and [salt] and [acid] represent the concentrations of the salt and the acid, respectively.

The buffer capacity depends on the concentrations of both the acid and the salt. The greater the concentrations of both the acid and its conjugate base (salt) in the solution, the greater the buffering capacity.

In this case, we need to compare the concentrations of HF (acid) and NaF (salt) in the given solutions. Let's evaluate each option:

A) 0.821 M HF and 0.909 M NaF
B) They are all buffer solutions and would all have the same capacity.
C) 0.100 M HF and 0.217 M NaF
D) 0.121 M HF and 0.667 M NaF
E) 0.821 M HF and 0.217 M NaF

By comparing the concentrations of both HF and NaF in each option, we can see that option D) 0.121 M HF and 0.667 M NaF has the highest concentrations of both the acid and salt. Therefore, option D) has the greatest buffering capacity among the given solutions.

Remember, the buffer capacity is maximized when the concentrations of both the acid and its conjugate base are high.

To determine which solution has the greatest buffering capacity, you need to compare the concentrations of the weak acid and its conjugate base in each solution. The greater the concentration of both the weak acid and its conjugate base, the greater the buffering capacity.

Let's analyze the concentration of the weak acid (HF) and its conjugate base (NaF) for each solution:

A) 0.821 M HF and 0.909 M NaF
B) They are all buffer solutions and would all have the same capacity.
C) 0.100 M HF and 0.217 M NaF
D) 0.121 M HF and 0.667 M NaF
E) 0.821 M HF and 0.217 M NaF

Comparing the concentrations:

A) Concentration of HF: 0.821 M
Concentration of NaF: 0.909 M

B) Since all the solutions are buffer solutions, they would all have the same capacity.

C) Concentration of HF: 0.100 M
Concentration of NaF: 0.217 M

D) Concentration of HF: 0.121 M
Concentration of NaF: 0.667 M

E) Concentration of HF: 0.821 M
Concentration of NaF: 0.217 M

From the given options, the solution with the greatest buffering capacity would be the one with the highest concentrations of both HF (weak acid) and NaF (conjugate base). Comparing the concentrations, we can see that option D) 0.121 M HF and 0.667 M NaF has the highest concentrations of both components, indicating that it would have the greatest buffering capacity.

Though studying on Thanksgiving can be difficult, it shows dedication and determination. Wishing you a successful study session! Happy Thanksgiving!

I will do the first one and you can do the others.

pH = pKa + log[(base)/(acid)]
I found Ka of 7.2E-4 for HF but you need to use the number in your text. pKa from this value is 3.14.
pH = 3.14 + log(NaF/HF)
pH = 3.14 + log(0.4545/0.4105) = 3.18
[Note: I have divided the molarity by 2 to find moles in 0.5L of each since the buffer capacity is defined as the moles of base or acid that can be added to a LITER of buffered solution without changing the pH more than +/- 1. The answer to your question is to determine how much base (or acid) that can be added to each without changing the pH more than +/- 1; i.e., between range 4.18 to 2.18)

......HF +..... NaOH ==> NaF +... H2O
start 0.4105... 0........0.4545...0
change -x......+x........+x.......+x
end...0.4105-x..0........0.4545+x....N/A

4.18 = 3.14 + log(0.4545+x/0.4105-x)
1.04 = log etc.
10.96 = (0.4545+x)/(0.4105-x)
solve for x and I obtained
0.338 for moles NaOH I can add.
If you chose to go the other way (adding acid) the answer will be slightly different. I found 0.369 moles acid which is why some profs don't like to use this definitions because +/- 1 pH unit isn't the same for base and acid. At any rate, compare the numbers.