Question

How many moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 3.50? (Ka for HF is 6.8 x 10-4)

Answer 1

We have relation, pKa = - log Ka

$\therefore$ pKa = - log ( 6.8 $\times$ 10 ^-04 ) = 3.17

We have, No. of moles = Molarity $\times$ volume of solution in L

$\therefore$ No. of moles of HF = 0.200 mol / L $\times$ 0.2000 L = 0.0400 mol

Consider a reaction , HF + NaOH $\rightarrow$ NaF + H₂O

Let's use ICE table.

Concentration ( moles)	HF	NaOH	NaF
Initial	0.0400	X
Change	- X	-X	+X
Equilibrium	0.0400	0.0	X

When NaOH is added to HF solution, some of the HF molecules are converted into NaF. Hence, at this stage solution contain weak acid HF and it's salt NaF. This solution acts as a buffer solution and it's pH is calculated by using Henderson's equation.

pH = pKa + log [NaF] / [HF]

3.50 = 3.17 + log X / 0.0400 - X

3.50 - 3.17 = log X / 0.0400 - X

0.33 = log X / 0.0400 - X

X / 0.0400 - X = 2.14

X = 2.14 ( 0.0400 - X)

X = 0.0856 - 2.14 X

X+2.14 X = 0.0856

3.14 X = 0.0856

X = 0.0856 / 3.14 = 0.0273 moles.

ANSWER : Moles of NaOH added to 200 ml 0.2 M HF solution to produce solution of pH = 3.50 is 0.0273.