Question

You are asked to prepare 2.6 L of a HCN/NaCN buffer that has a pH of 9.84 and an osmotic pressure of 1.73 atm at 298 K. What masses of HCN and NaCN should you use to prepare the buffer? (Assume complete dissociation of NaCN.)

Answer 1

For a buffer problem like this, it is best to use the Henderson-Hasselbalch equation:
pH = pKa + log (base/acid)

The Ka of hydrocyanic acid is 6.2 x 10-10 (check your textbook's appendix, since Ka values sometimes vary among textbooks), so the pKa of HCN is 9.207.

We now must determine the ratio of base to acid (in moles/liter)

9.84 = 9.207 + log (base/acid)

0.633 = log (base/acid)   --- now we must get rid of the log on the right by raising a base of 10 to both sides, since 10log x = x, by definition.


4.295 = base/acid, or NaCN/HCN

Before we continuue on, let's determine how many moles of HCN and NaCN are present in the buffer by using the ideal gas law, VP = nRT, or n = VP/RT

(2.6L)(1.73 atm)/(0.082057 atm L mol-1 K-1)(298 K) = 0.0614 moles present

The moles present represent the sums of the moles of HCN and NaCN not simply one of them. We therefore need algebra to determine the masses present.

Using 4.295 = NaCN/HCN and 0.0614 moles of NaCN + HCN, we can do this:

(1) 4.295(HCN) = NaCN

(2) HCN = 0.0614 moles - NaCN

Plug 2 into 1: 4.295 (0.0614 moles - NaCN) = NaCN

0.263713 - 4.295 NaCN = NaCN

0.263713 = 5.295 NaCN

0.0498 moles of NaCN

Use this value in (2): HCN = 0.0614 - 0.0498 = 0.0116 HCN

Our moles of NaCN and HCN are now known:

0.0116 HCN, 0.0498 NaCN

Now convert these values into grams:

0.0116 mol HCN (27.02534 g/mol) = 0.3135 g HCN, or 0.3135 g

0.0498*49.0072 g NaCN, or 2.44 g.