Question

PART A:

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

0.260M

4.70×10⁻² M

2.30×10⁻² M

In which cases can you not make the simplifying assumption that x is small? (choose one of the below options)

only in (a)

only in (b)

in (a) and (b)

in (b) and (c)

PART B:

A 8.0×10⁻² M solution of a monoprotic acid has a percent dissociation of 0.60%. Determine the acid ionization constant (Ka) for the acid.

Answer 1

A.

HF is a weak acid so you will have to make an ICE table

HF + H2O ----> H3O+ + F-
0.260
-x........................+x.....+x
0.260-x...............x........x

x^2/(0.260-x) = 3.5x10^-4

Solve for X

Assume 0.260-x ~ 0.260

-log[X] to find pH

= 1.73

so 1.73 / 0.260 is 6.6 % and less than 5%, then your assumption is not good. Calculate same way for remaining two.

Part B-

Please check my calculation, I am very weak mathematician.

% ionization = [H+]/[acid overall] x 100

Therefore [H+] = %ionization/100 x [acid overall]

[H+] = 0.60/100 x 0.080
[H+] = 0.00048M

HA <--> H+ + A-

Concentration [HA] = 0.080M - x <--- this x is insignificant cause percent ionization is so low.
Concentration [H+] = 0.00048M
Concentration [A-] = 0.00048M

Ka = [products]/[reactants]

Ka = (0.00048)^2/0.080 = 2.88x10^-6