Question

You are asked to prepare 1.1 L of a HCN/NaCN buffer that has a pH of 9.63 and an osmotic pressure of 1.96 atm at 298 K.

What masses of HCN and NaCN should you use to prepare the buffer? (Assume complete dissociation of NaCN.)

Answer 1

Osmotic pressure is given by

= MRT

Where, = 1.96 atm

R= 0.0821 L.atm /mol.K

T = 298 K

so, 1.96 atm = ( 0.0821 L.atm /mol.K * 298 K) M

M = 0.0801

molarity = moles / Volume of solution = 0.0801

Moles of HCN and NaCN in buffer = 0.0801*1.1L = 0.0881 moles

HCN + NaCN = 0.0881......(1)

Let moles of HCN be x

Then the moles of NaCN = 0.0881 - x

From literature, pKa = 9.21

From Henderson - Hasselbalch equation

pH = pKa + log [NaCN]/[HCN]

9.63 = 9.21 + log [NaCN]/[HCN]

[NaCN]/[HCN] = 2.63

[NaCN] = 2.63[HCN]

From (1)

HCN + 2.63[HCN] = 0.0881

3.63[HCN] = 0.0881

[HCN] = 0.0243 moles

Thus, moles of NaCN = 0.0881 - 0.0243 = 0.0638 moles

Mass of HCN = moles * molar mass = 0.0243 mol * (27.02534 g/mol) = 0.657 g

Mass of NaCN = Moles * molar mass = 0.0638 moles *49.0072 g/mol = 3.13 g

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