buffer preperation

Question

1) How many moles of sodium hydroxide would have to be added to 125 mL of a 0.343 M hydrocyanic acid solution, in order to prepare a buffer with a pH of 9.390?

=moles

Answer 1

We will need the Ka for HCN, which I found to be (Ka of HCN is 4.9 × 10^-10

Using the Henderson Hasselbalch equation, pH = pKa + log [salt]/[acid], we can find the ratio [salt]/[acid].

pKa = -log Ka = 9.31

9.390 = 9.31 + log [salt]/[acid],

0.08 = log [salt]/[acid],

[salt]/[acid] = 1.20

Initial moles HCN = 0.125 L x 0.343 mol/L = 0.0429 moles

Let x = moles of NaCN formed...

x / 0.0429-x = 1.2

x = 0.0234 moles NaCN needs to be formed

Since 1 mol HA reacts with 1 mole NaOH to produce 1 mol NaCN, 0.0234 moles NaOH need to be added.

Check of calculations:

HA + NaOH =====> NaCN + H2O

0.0429.....x..........................0................Initial

-x...........-x..........................+x...............Change

0.0429-x...0........................x.................Equilibrium

If x = 0.0234 (as calculated above), what will the NaCN/HCN ratio be?

0.0429 - 0.0234 = 0.0195 = HA

NaCN = 0.0234

Ratio = 0.0234 / 0.0195 = 1.20 This is what was calculated above.