Question

A 0.2000M solution of hydroflouric acid is only partially ionized. Using measurements of the pH of the solution, [H+] was determined to be 3.184 × 10-2 M. Calculate the acid dissociation constant of hydroflouric acid.a. 7.023 × 10-3b. 5.534 × 10-3c. 6.669 × 10-3d. 6.029 × 10-3

Answer 1

a. 7.023 × 10⁻³

Step-by-step explanation

Given:

Initial [HF] = 0.2000 M

[H⁺] = 3.8184 x10⁻² M

The dissociate of HF will give:

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

Therefore the dissociation constant of HF is:

`K_c=([H^+][F^-])/([HF])`

Putting [H⁺] = [F⁻] = 3.8184 x10⁻² M and [HF] = 0.2000 M

`\begin{gathered} K_c=(3.8184*10^(-2)*3.8184*10^(-2))/(0.2000) \\ \\ K_c=7.023*10^(-3) \end{gathered}`