You are asked to prepare 3.4 L of a HCN/NaCNbuffer that has a pH of 9.69 and an

Question

You are asked to prepare 3.4 L of a HCN/NaCNbuffer that has a pH of 9.69 and an osmotic pressure of 1.59 atm at 298 K. What masses of HCN and NaCN should you use to prepare the buffer? (Assume complete dissociation of NaCN.)

I have attempted this multiple times and I have yet to obtain the correct answer. My answers have been: 1.5 grams of HCN and 8.2 grams of NaCN, .74 grams of HCN and 4.1 grams of NaCN, 1.5 grams of HCN and 8.1 grams of NaCN, .85 grams of HCN and 4.6 grams of NaCN, .50 grams of HCN and 2.7 grams of HCN.

None of these have been deemed correct. Any help is greatly appreciated.

Explanation / Answer

pKa for HCN = 9.31

pH = pKa + log { [salt] / [acid] }

9.69 = 9.31 + log { [salt] / [acid] }

log { [salt] / [acid] }= 9.69 – 9.31

log { [salt] / [acid] }= 0.38

[salt] / [acid] = 100.38

[salt] / [acid] = 2.40

[salt] = 2.40 [acid]

= i M R T

1.59 = (2) (M) (0.082) (298)

M = 1.59 / { (2) (0.082) (298) }

M = 0.0325 M

So, the molarity of the buffer is 0.0325 M

[salt] + [acid] = 0.0325

2.40 [acid] + [acid] = 0.0325

3.40 [acid] = 0.0325

[acid] = 0.0325 / 3.40 = 0.0096 M

Now,

[salt] + [acid] = 0.0325

[salt] = 0.0325 - [acid]

[salt] = 0.0325 – 0.0096

[salt] = 0.0229 M

Volume = 3.4 L

Moles of NaCN = 0.0229 M x 3.4 L = 0.07786 moles

Moles of HCN = 0.0096 M x 3.4 L = 0.03264 moles

Molar mass of HCN = 27.03 g/mol

So, 1 mole of HCN = 27.03 g

0.03264 moles of HCN = 0.03264 x 27.03 g = 0.88 g

Molar mass of NaCN = 49.01 g/mol

So, 1 mole of NaCN = 49.01 g

0.07786 moles of NaCN = 0.07786 x 49.01 g = 3.82 g

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