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.
3 answers
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.
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.
2 answers