Question

The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

6.0 × 10-5
3.5 × 10-4
1.1 × 10-9
2.0 × 10-9
none of the above

Answer 1

The value of Ka for the given hydrofluoric acid solution is 6.6 x 10^(-4).

Step-by-step explanation:

To calculate the value of Ka for hydrofluoric acid (HF), we need to use the equation for the ionization of HF in water:

HF(aq) + H2O(l) → H3O+(aq) + F-(aq)

We also know that the pH of the solution is 2.03. The pH is related to the concentration of H3O+ by the equation pH = -log[H3O+]. Hence, [H3O+] = 10^(-pH) = 10^(-2.03).

Next, we write an expression for the equilibrium constant (Ka) using the concentrations of the ions:

Ka = [H3O+][F-]/[HF]

Substituting the known values, we have Ka = (10^(-2.03))(10^(-2.03))/0.25 = 6.6 x 10^(-4).

Therefore, the correct answer is 6.6 x 10^(-4).

Answer 2

Answer : The correct option is,
`3.5* 10^(-4)`

Solution : Given,

pH = 2.03

Concentration of HF = 0.25 M

First we have to calculate the concentration of
`H^+` ion.

`pH=-\log [H^+]`

`2.03=-\log [H^+]`

`[H^+]=9.3* 10^(-3)M`

Now we have to calculate the value of
`K_a` for HF.

The equilibrium reaction will be

`HF\rightleftharpoons H^++F^-`

Concentration of
`H^+` = Concentration of
`F^-` =
`9.3* 10^(-3)M`

The expression for
`K_a` for HF will be,

`K_a=([H^+][F^-])/([HF])=((9.3* 10^(-3))* (9.3* 10^(-3)))/(0.25)=3.5* 10^(-4)`

Therefore, the value of
`K_a` for HF is,
`3.5* 10^(-4)`