Question

Determine the pH of an HF solution of each of the following concentrations.

0.290 M

4.50x10^-2 M

2.50x10^-2 M

Answer 1

Given :[HF]0.290M

HF is a weak acid that is hydrolysed in solution as follows:

HF +H2O <---->F- +H3O+

ka6.6*10^-4[F-][H3O+]/[HF]

ICE table

	[HF]	[F-]	[H3O+]
initial	0.290M	0	0
change	-x	+x	+x
equilibrium	0.290-x	x	x

6.6*10^-4[F-][H3O+]/[HF]x^2/(0.290-x)

As ka is small ,so very less dissociation takes place,thus 0.290>>x,

6.6*10^-4x^2/(0.290)

x[H3O+]0.0138M

pH-log [H3O+]-log(0.0138)1.9

pH1.9

2)

HF +H2O <---->F- +H3O+

ka6.6*10^-4[F-][H3O+]/[HF]

ICE table

	[HF]	[F-]	[H3O+]
initial	4.50*10^-2M	0	0
change	-x	+x	+x
equilibrium	4.50*10^-2-x	x	x

6.6*10^-4[F-][H3O+]/[HF]x^2/(4.50*10^-2-x)

As ka is small ,so very less dissociation takes place,thus 4.50*10^-2>>x,

6.6*10^-4x^2/(4.50*10^-2)

x[H3O+]5.45*10^-3M

pH-log [H3O+]-log(5.45*10^-3)2.3

pH2.3

3)

HF +H2O <---->F- +H3O+

ka6.6*10^-4[F-][H3O+]/[HF]

ICE table

	[HF]	[F-]	[H3O+]
initial	2.5*10^-2M	0	0
change	-x	+x	+x
equilibrium	2.5*10^-2-x	x	x

6.6*10^-4[F-][H3O+]/[HF]x^2/(2.5*10^-2-x)

As ka is small ,so very less dissociation takes place,thus 2.5*10^-2>>x,

6.6*10^-4x^2/(2.5*10^-2)

x[H3O+]4.062*10^-3M

pH-log [H3O+]-log(4.062*10^-3)2.4

pH2.4