Answer to Question #92443 in Chemistry for Tj

Question

A 0.20 M solution of hydrofluoric acid (HF(aq)) has a pH of 3.96 at 25 oC. Calculate the Ka and Percent Ionization of hydrofluoric acid.

HF(aq) + H2O(l)F-(aq) + H3O+(aq)

Answer 1

Conditions of question: reaction HF(aq) + H₂O(l) = F-(aq) + H₃O+(aq)

p-H (solution of hydrofluoric acid) = 3.96 at 25^oC

С(HF(aq))=0.20 M

To find: the Ka and Percent Ionization ( α )of hydrofluoric acid  

Solution:  The expression of the acidity constant (Ka) for hydrofluoric acid has the form

$Ka=([F-]*[H30+])/[HF]$

$pH=-log[H3O+]$ , $pKa=-logKa$

Transforming, we get the Henderson-Hasselbach equation:

$-pKa=-pH+log([F-]/[HF])$ or

$pH=pKa+log([F-]/[HF])$.

As stated above, pH=-log[H3O+]. So, [H₃O+]=10^-3.96=1.097*10^-4.

According to the acid dissociation equation [H₃O+]=[F-]=1.097*10^-4.

Substitute data and get 3,96=pKa + log((1.097*10^-4)/0.20)

pKa=3.96-(-3.26)=7.22

Ka=10^-7.22=6.062*10^-8

Percent Ionization α represents the fraction of disintegrated molecules. If the total concentration of electrolyte (C) is multiplied by this fraction, then we obtain the concentration of any of the ions. So, $Ka=C* α *C* α /(C-C* α )$

$α =$ (Ka/C)^0.5 = ((6.062*10^-8)/0.20)^0.5 = 5.489*10^-4 = 0.055 %

Answer:  Ka=6.062*10^-8, α = 0.055 %.