Four sets of conjugate acid/base pairs and Ka of the acid are given. Which pair would be the best to form a buffer at pH = 9.2? You can use the Henderson-Hasselbalch equation to answer this question.

HCNO/CNO- 3.5 x 10-4

HC2O4-/C2O4- 5.1 x 10-5

H2CO3/HCO3- 4.3 x 10-7

HCN/CN- 4.9 x 10-10

I know that you have to use the Henderson-Hasselbalch equation and I know what the equation is. I just have no idea how to set it up with what I've been given.

Actually you don't need to use the HH equation at all. The rule is that to make a buffer you want pKa to be within about 1 unit of the desired pH. So convert each of the Ka values to pKa values and choose the one that is within about 1 unit. And if you are good at scanning data, you KNOW the pKa for the first one is 3+, the second is 4+, the third is 6+ and the fourth is 9+. So you convert the last one and see if that meets the criteria. I suggest you convert all of them and see how easy it is to do.

I took it as 4.9 x 10^-10 was correct from what you said and when I chose that as an answer it said it was wrong?

You took my response correctly and the correct answer is the HCN/CN pair because if Ka = 4.9E-10 then pKa = 9.31 which is quite close to the desired pH of 9.2; therefore, it is easy to make solutions in which the ratio of base/acid is close to 1.

If you use the HH equation, for example we can do #1, it is done this way.
The problem CLEARLY says that the numbers given are the Ka values; therefore, pKa for Ka of 3.5E-4 is 3.45
Substitute that into the HH equation and you get
pH = pKa + log(base)/(acid)
9.2 = 3.45 + log base/acid
Then (base)/(acid) = b/a = 5.62E5
That means the (CNO^-)/(HCNO) ratio would have to be 5.62E5. That means if (HCNO) = 1M then (CNO^-) would be 5.62E5M which is essentially impossible to get. I ran through the other numbers and came up with these values for base/acid.
For Ka 3.5E-4 pKa = 3.45 b/a =5.62E5
For Ka 5.1E-5 pKa = 4.29 b/a =8.13E4
For Ka 4.3E-7 pKa = 6.37 b/a =6.76E2
For Ka 4.9E-10 pKa = 9.31 b/a =0.776

You should confirm those numbers but the point is that it is easy to make a solution in which the ratio is very close to 1. The further away from 1 the harder (up to impossible) it is to do. I'm sticking by my answer. I thought perhaps the author of the problem had made the acid/base ratio with different Ka values; however, I look the Ka values up on the web. Although some were not exactly the same, they were close enough. The author did NOT take the reciprocal which is what I feared. The only other thing I can suggest is that if you input 4.9E-10 it may have said you were wrong because the question asks for the acid/base pair. So the answer is HCN/CN or d or however you are to enter the answer.

To determine which pair would be the best to form a buffer at pH = 9.2, we can use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the ratio of the concentrations of the conjugate acid and base in a buffer.

The Henderson-Hasselbalch equation is given by:

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

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

In this case, we are given the pKa values for the acids in each of the given pairs. To determine the ratio of [A-] to [HA], we need to calculate the concentrations of the conjugate base and acid.

Let's calculate the concentrations for each pair:

1. HCNO/CNO-:
Given: pKa = 3.5 x 10^(-4)
Since pKa = -log(Ka), we can find Ka by taking the antilog:
Ka = 10^(-pKa) = 10^(-3.5 x 10^(-4)) ≈ 0.876
Since we are given the Ka value, we can assume that the concentration of HCNO and CNO- are the same, for example, let's assume it to be x.

[H+] = 10^(-pH) = 10^(-9.2) ≈ 6.31 x 10^(-10)

Using the Henderson-Hasselbalch equation:

9.2 = -log(0.876) + log(x/x)

Simplifying:

9.2 = 3.5 x 10^(-4) + 0

The concentration of HCNO and CNO- does not affect the pH of the solution. Therefore, this pair is not suitable for forming a buffer at pH = 9.2.

2. HC2O4-/C2O4-:
Given: pKa = 5.1 x 10^(-5)
Using the same approach as above, let's assume the concentration of HC2O4- and C2O4- to be x.

[H+] = 10^(-pH) = 10^(-9.2) ≈ 6.31 x 10^(-10)

Using the Henderson-Hasselbalch equation:

9.2 = -log(5.1 x 10^(-5)) + log(x/x)

Simplifying:

9.2 = 5.1 x 10^(-5) + 0

The concentration of HC2O4- and C2O4- does not affect the pH of the solution. Therefore, this pair is not suitable for forming a buffer at pH = 9.2.

3. H2CO3/HCO3-:
Given: pKa = 4.3 x 10^(-7)
Using the same approach as above, let's assume the concentration of H2CO3 and HCO3- to be x.

[H+] = 10^(-pH) = 10^(-9.2) ≈ 6.31 x 10^(-10)

Using the Henderson-Hasselbalch equation:

9.2 = -log(4.3 x 10^(-7)) + log(x/x)

Simplifying:

9.2 ≈ 6.37 + 0

The concentrations of H2CO3 and HCO3- do affect the pH of the solution. Therefore, this pair would be suitable for forming a buffer at pH = 9.2.

4. HCN/CN-:
Given: pKa = 4.9 x 10^(-10)
Using the same approach as above, let's assume the concentration of HCN and CN- to be x.

[H+] = 10^(-pH) = 10^(-9.2) ≈ 6.31 x 10^(-10)

Using the Henderson-Hasselbalch equation:

9.2 = -log(4.9 x 10^(-10)) + log(x/x)

Simplifying:

9.2 ≈ 9.31 + 0

The concentrations of HCN and CN- do affect the pH of the solution. Therefore, this pair would be suitable for forming a buffer at pH = 9.2.

Based on the calculations, the pair HCN/CN- with a Ka value of 4.9 x 10^(-10) would be the best to form a buffer at pH = 9.2.