Question

1. HF (aq) = H+ (aq) + F- (aq)

K 6.8 x 10-4
-
II. H₂C₂O4 (aq) = 2H+ (aq) + C₂0²-
K = 3.8 x 10-6
What is the K value for the
reaction below?
2HF (aq) + C₂02 (aq) =
2F (aq) + H₂C₂O4 (aq)
[?]
K = [?] x 10¹

Answer 1

In the image above/below (i cant tell), you want to try and replicate the third reaction given the first two reaction, so you just have to manipulate the reactants/products to cancel out and result in the reaction we’re looking for.

Remember:
when you double moles: square the equilibrium constant

when you half the moles: Square root the equilibrium constant

when you reverse a reaction: take the reciprocal of the equilibrium constant