a. The dissociation constant of the weak acid HF is K₂ = [H][F]. If at equilibrium
Ka
[HF]
the concentrations of H* and F are each 2.6 x 10-3 M and the concentration of
HF is 0.010 M, what is K₂ for HF? (3 points)

This is a bit of a confusing question, since it incorrectly states that the acid disassociation constant is simply [H+][F-].
Ka = [H+][F-]/[HF] , so plugging in the values, we get
[tex]K=\frac{[2.6*10^{-3}][2.6*10^{-3}]}{[0.01]}\\K=\frac{[2.6*10^{-3}]^2}{[0.01]}\\K=0.000676=6.76*10^{-4}[/tex]

If we were to use the equation in the question, Ka=[H+][F-], we would get the answer 6.76 * 10^-6 , which is not the Ka for HF

a. The dissociation constant of the weak acid HF is K₂ = [H][F]. If at equilibrium Ka [HF] the concentrations of H* and F are each 2.6 x 10-3 M and the concentration of HF is 0.010 M, what is K₂ for HF? (3 points)

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a. The dissociation constant of the weak acid HF is K₂ = [H][F]. If at equilibrium
Ka
[HF]
the concentrations of H* and F are each 2.6 x 10-3 M and the concentration of
HF is 0.010 M, what is K₂ for HF? (3 points)