Question

You are asked to prepare 3.1 L of a HCN/NaCN buffer that has a pH of 9.59 and an osmotic pressure of 1.98 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.

mHCN, mNaCN =

Answer 1

Answer : mHCN , mNaCN = 0.76 , 2.56 respectively

Solution :

Given, Osmotic pressure of the buffer HCN/NaCN =1.98

Temperature = 298K

Volume of solution = 3.1L

This value helps us to find total concentration of the buffer and hence , total number of moles (NaCN +HCN)

Using,  π = iMRT ...............................(a)

π = osmotic pressure

M= molarity

R = gas constant = 0.082 L atm K⁻¹ mol^−1.

T = temperature

i = van't Hoff factor

i = 3 because NaCN gives Na+ and CN-, and HCN give H+ and CN-. So you have H+, Na+ and CN- an so total absolute charge , i = 3

Using (a)

1.98 = 3 * M * 0.082 * 298

M = 0.027

Molarity = Moles / Volume

Moles (NaCN +HCN) = 0.027 * 3.1 = 0.0837

Now, we will find the ratio moles of NaCN/ HCN using  the Henderson Hasselbalch equation

pH = pKa + log [NaCN]/[HCN] ..................(b)

pKa of HCN = 9.31

pH of buffer = 9.59

Uisng (b)

9.59 = 9.31 + log [NaCN]/[HCN]

log [NaCN]/[HCN] = 9.59 - 9.31

log [NaCN]/[HCN] = 0.28

[NaCN]/[HCN] = 10 ^0.28

[NaCN]/[HCN] = 1.905

We have calculated the ratio for the concentration of NaCN and HCN. Since , the volume of the solution is same ,so, the ratio of moles (n) of NaCN and HCN is same as 1.905

n _NaCN / n _HCN = 1.905

From above relation , n _NaCN = 1.905 n _HCN ..............(c)

Also, n _NaCN + n _HCN = 0.0837

1.905 n _HCN + n _HCN = 0.0837

2.905 n _HCN = 0.0837

n _HCN = 0.0837 / 2.905

  n _HCN = 0.028

and from (c) n _NaCN = 0.05334

Mass of HCN = Moles * molar mass   Mass of NaCN = Moles * molar mass

= 0.028 * 27.025 and = 0.053 * 49.007

= 0.756 = 2.597

mHCN = 0.76 and mNaCN =2.56 (rounded off to 2 decimal places)