In: Chemistry
Part A
If a solution of HF (Ka=6.8×10−4) has a pH of 2.80, calculate the concentration of hydrofluoric acid.
Express your answer using two significant figures.
n = mol
Sol:-
Let concentration of HF = X mol/L
Given pH = 2.80
[H+] = 10-pH
= 10-2.80
= 0.00158 mol/L
ICE table of HF is :
............................................HF (aq) <---------------------------> H+ (aq)........................+............................F- (aq)
Initial (I)..............................X mol/L.........................................0.0 mol/L....................................................0.0 mol/L
Change (C).......................-0.00158 mol/L............................+0.00158 mol/L..........................................+0.00158 mol/L
Equilibrium (E)..............(X-0.00158) mol/L...............................0.00158 mol/L...........................................0.00158 mol/L
Acid dissociation constant (ka) is equal to the ratio of product of the molar concentration of products to the molar concentration of reactants raise to power of stoichiometric coefficient at equilibrium stage of the reaction.
Expression of ka is :
ka = [H+].[F-] / [HF]
6.8 x 10-4 = (0.00158)2 / (X-0.00158)
6.8 x 10-4 X - 1.0744 x 10-6 = 2.4964 x 10-6
6.8 x 10-4 X = 2.4964 x 10-6 + 1.0744 x 10-6
6.8 x 10-4 X = 3.5708 x 10-6
X = 3.5708 x 10-6 / 6.8 x 10-4
X = 5.25 x 10-3
Hence, concentration of HF = 5.25 x 10-3 M |