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Hello everyone, I'm back again with another question, this time about buffer solutions. This assignment is divided into two parts, I've done the first one and I am quite confident in my answers. As for the second one, I can't see where the problem is in my calculations, but the results do not seem right to me.

Here is the assignment: the objective is to make a buffer solution of 1L with pH = 8,50. Starting from 0,01mol KCN, how many moles of HCl need to be added in order to obtain the desired buffer solution, knowing that Ka of HCN is 4,8 x 10-10?

What will be the variation of pH after adding 0,00005 moles of HClO4 to 100ml of buffer solution?

Now as I said I think I have understood the first part. The problem with the second question is that the variation of pH I obtained is 3,12, but to me that seems wrong as, considering we are talking about buffer solution, the variation should be minimal. What have I done wrong?

Here's my attempt: 

Molarity KCN = moles KCN / Volume

Molarity KCN = 0,01mol/1L = 0,01mol/L (which from now on I'll call Cb, concentration of the base)

I then calculated pKa for HCN:

pKa HCN = -logKa

pKa HCN = -log (4.8 x 10-10)

pKa HCN = 9.32

I then used Henderson-Hasselbach equation:

pH = pKa + log [base]/[acid] (or pH = pKa + log Cb/Ca)

And I modified the equation to find log[base]/[acid]:

log[base]/[acid] = pH-pKa

log[base]/[acid] = 8,50 - 9,32 = -0,82

[base]/[acid] = 10-0,82 = 0,15

Ca = Cb/0,15

Ca = 0,01mol/L/0,15 = 0,06mol/L

number of moles of HCl = 0,06mol/L x 1L = 0,06mol. This is where the first part ends and the second begins

Molarity HClO4 ( = Ca) = 0,00005mol/0,1L = 0,0005mol/L

Molarity KCN = 0,01mol/0,1L = 0,1mol/L

I then used Henderson-Hasselbach again:

pH = pKa + log[base]/[acid]

pH2 = 9,32 + log(0,1mol/L / 0,0005mol/L)

pH2 = 9,32 + 2,30 = 11,62

I finally calculated delta pH:

pH2 - pH1 = 11,62 - 8,50 = 3,12, which I think is wrong because the variation is too great for a buffer solution. Any ideas as to where I went wrong??

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