A 0.400 M solution of an acid, HA, has a pH = 1.301. What is the value of the ionization constant, Ka, for?

Question

A 0.400 M solution of an acid, HA, has a pH = 1.301. What is the value of the ionization constant, Ka, for
this acid?

A) 5.00 x 10-2 B) 1.25 x 10-3 C) 5.56 x 10-3 D) 6.25 x 10-3

Answer 1

The way to solve such problems is

.. .. .. .. .. HA <=> H(+) +A(-)
Initial .. .. .C
Dissoc. .. x
Produce .. .. .. .. .. x .. .. ..x
At Equil. .C-x .. .. x .. .. .. x

Ka= [H+][A-]/[HA]= x^2/(C-x)
but x=[H+] and pH=-log[H+]= -logx => x=10^-pH= 10^-1.301= 0.05

Ka=(0.05)^2 / (0.4-0.05)= 7.14*10^-3

Sometimes, when you are solving for x, you don't want to bother solving the quadratic so you do the assumption that C >> x and thus C-x=C simplifying the equation to

Ka=x^2/C

If you used this simplified equation then

Ka=(0.05)^2/0.4= 6.25*10^-3 which is answer D.

However when we do this simplification we always check if the calculated value for x is truly <
You should point this out to your teacher.

Answer 2

The acid HA dissociates: HA ? H+ + A- Now do not ignore that our purpose is to remedy for [H+] you should understand the Ka equation: Ka = [H+] [A-] / [HA] Chemist make their lives elementary with techniques from affirming 2 issues: a million) you will discover from the dissociation equation that [H+] = [A-] - if you must write down [H+]² for the numerator 2) [HA] interior the equation is the most objective of the unionised acid. yet - because the ionisation is amazingly low , we may be able to ignore about this and use the wanted concentration for [HA] Now rewrite the Ka equation with numbers: 4.24*10^-6 = [H+]² / 0.616 [H+² = (4.24*10^-6) * 0.616 [H+]² = 2.612*10^-6 [H+] = ?( 2.612*10^-6) [H+] = a million.616*10^-3M Now you've the molar concentration of H+ in answer. Calculate the % ionisation: % ionisation = (a million.616*10^-3) / 0.616 * one hundred % ionisation = 0.262%

Answer 3

C - definitely C