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UkoKoshka [18]
3 years ago
8

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?

Chemistry
2 answers:
Mazyrski [523]3 years ago
8 0
Last option that is none of above is right answer.
MrRa [10]3 years ago
7 0

Answer : The correct option is, 3.5\times 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\times 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\times 10^{-3}M

The expression for K_a for HF will be,

K_a=\frac{[H^+][F^-]}{[HF]}=\frac{(9.3\times 10^{-3})\times (9.3\times 10^{-3})}{0.25}=3.5\times 10^{-4}

Therefore, the value of K_a for HF is, 3.5\times 10^{-4}

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