Question

You are asked to prepare 1.6 L of a HCN/NaCN buffer that has a pH of 9.90 and an osmotic pressure of 1.12 atm at 298 K.

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

Express your answers using two significant figures separated by a comma.

Answer 1

Sol:-

We know

pi = CRT .............(1)

here, pi = Osmotic pressure = 1.12 atm ( given)

R = universal gas constant = 0.0821 L atm K^-1 mol^-1

T = Temperature = 298 K (given)

C = Concentration of the solution = ?

From equation (1), we have

C = pi / RT

C = 1.12 atm / 0.0821 L atm K^-1 mol^-1 x 298 K

C = 0.0458 mol L^-1

also

Molarity i.e Concentration = Number of moles / volume of the solution in L

Given Volume = 1.6 L

so

Total number of moles ( i.e HCN + NaCN ) = 0.0458 mol L^-1 x 1.6 L = 0.07328 mol

Let number of moles of HCN = n

so

Number of moles of NaCN = (0.07328 - n ) moles

According to Henderson-Hasselbalch equation for acidic buffer solution i.e

pH = pK_a + log [ Conjugate base ] / [ Acid]

pH = pK_a + log [ NaCN ] / [ HCN] ........(2)

note that pK_a of HCN = 9.31

given pH = 9.90

so from equation (2), we have

9.90 = 9.31 + log ( 0.07328 - n / n )

log ( 0.07328 - n / n ) = 9.90 - 9.31

log ( 0.07328 - n / n ) = 0.59

0.07328 - n / n = 10^0.59

0.07328 - n / n = 3.89

0.07328 - n = 3.89n

3.89n + n = 0.07328

4.89n = 0.07328

n = 0.07328 / 4.89

n = 0.014985685 mol

so

number of moles of HCN = n = 0.014985685 mol

and

Number of moles of NaCN = (0.07328 - n ) = 0.07328 - 0.014985685 = 0.058294315 mol

we know

Number of moles = mass of substance in gram / Gram molar mass of substance

so

Mass of HCN = Number of moles of HCN x Gram molar mass of HCN

Mass of HCN = 0.014985685 mol x 27.0253 g/mo = 0.40 g

Hence mass of HCN = 0.40 g

similarly

Mass of NaCN = Number of moles of NaCN x Gram molar mass of NaCN

Mass of NaCN = 0.058294315 mol x 49.0072 g/mol = 2.9 g

Hence mass of NaCN = 2.9 g