Question

How many mL of 6.0 NaOH must be added to 0.50 L of 0.20 M HNO2 to prepare a pH = 3.86 buffer?

Answer 1

Here we use the Henderson–Hasselbalch equation.

pH = pKa + log [A-]/[HA]

The reaction of HNO2 and NaOH is as follows

HNO2(aq) + NaOH(aq) ------> NaNO2(aq) + H2O(l)

Here, base NaNO2 is produced .

Now calculate the ratio [NaNO2]/[HNO2] as follows

pKa = -log Ka
pKa = -log (4.0 x 10-4) = 3.40

pH = pKa + log [NaNO2]/[HNO2]

3.86 = 3.40 + log [NaNO2]/[HNO2]

log [NaNO2]/[HNO2] = 3.86 - 3.40 = 0.46

[NaNO2]/[HNO2] = antilog 0.46 = 2.88

[NaNO2]/[HNO2] = 2.88

[NaNO2] = 2.88 [HNO2]

This means in the buffer the concentration of the salt [NaNO2] must be 2.88 times higher than concentration of the acid [HNO2].

As the volume is same for all concentrations then;

mol of NaNO2 = 2.88 x mol of HNO2

mol HNO2 = 0.50 L (0.20 mol/L) = 0.10 mol

Now, let x = volume of NaOH needed

therefore
mol NaOH = x L (6.0 mol/L) = 6x mol

As the reaction follows 1:1 mole ratio
mol NaNO2 produced = 6x mol

Hence we must have

mole HNO2 left = 0.1 - 6x

But, mol of NaNO2 = 2.88 x mol of HNO2

Then,

6x = 2.88 (0.1 - 6x)

Now solve x as follows:

6x + 17.28x = 0.288

23.28x = 0.288

x = 0.01237 L

= 12.37 mL or 12.0 ml