Question

You are asked to prepare 1.7 L of a HCN/NaCN buffer that has a pH of 9.66 and an osmotic pressure of 1.72 atm at 298 K.

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

Answer 1

P = MRT

M = P/(RT) = 1.72/(0.082*298) = 0.0703879522 M

then

pKa = HCN = 9.2

pH = pKa + log(NaCN/HCN)

9.66 = 9.20 + log(ratio)

ratio = 10^(9.66-9.20) = 2.88403

NACN = 2.88403*HCN

and we know that

mol = M*V = 1.7*0.0703879522 = 0.119659 mol (in total)

0.119659 = NaCN + HCN

NACN = 2.88403*HCN

from the last 2 equations:

0.119659  = 2.88403*HCN+ HCN

3.88403HCN = 0.119659

HCN = 0.119659 /3.88403

HCN = 0.0308 mol

then

0.119659  = NaCN + HCN

NACN = 0.119659 - 0.0308 =0.088859

mass of HCN = mol*MW = 0.0308 *27.0253 = 0.83237924 g of HCN

mass of NaCN = mol*MW = 0.088859*49.0072 = 4.354730 g of NACN