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.)

Answer 1

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] = 10^0.38

[salt] / [acid] = 2.40

[salt] = 2.40 [acid]

π = 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