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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

Respuesta :

Answer

a. 7.023 × 10⁻³

Explanation

Given:

Initial [HF] = 0.2000 M

[H⁺] = 3.8184 x10⁻² M

The dissociate of HF will give:

[tex]HF\rightarrow H^++F^-[/tex]

Therefore the dissociation constant of HF is:

[tex]K_c=\frac{[H^+][F^-]}{[HF]}[/tex]

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

[tex]\begin{gathered} K_c=\frac{3.8184\times10^{-2}\times3.8184\times10^{-2}}{0.2000} \\ \\ K_c=7.023\times10^{-3} \end{gathered}[/tex]