For the acid-ionization constant, there's a formula for that:
K[sub]a[/sub] = ([A[sup]-[/sup]][H[sup]+[/sup]]) / [HA]
where [A[sup]-[/sup]] is the concentration of the anion, [H[sup]+[/sup]] is the concentration of the proton, and [HA] is the concentration of the acid (all at equilibrium). Figuring out how much of the acid has dissociated should be easy to figure out:
3% of 0.010 is 0.00030, so the acid ionization constant equation for the first problem should be
K[sub]a[/sub] = [0.00030 M][sup]2[/sup]/(0.0097 M) = 9.3 x 10[sup]-6[/sup] M.[footnote]Wikipedia, Acid Dissociation Constant [http://en.wikipedia.org/wiki/Acid_dissociation_constant][/footnote]
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For the chemical equilibrium, there's also an equation for that (again, thanks, Wikipedia!):
For an chemical reaction aA + bB + ... --> ... yY + zZ, where the lower case letters are coefficients and the upper case letters are the reactants and products, the equilibrium constant K[sub]c[/sub] can be expressed as
K[sub]c[/sub] = (...[Y][sup]y[/sup][Z][sup]z[/sup]) / ([A][sup]a[/sup][sup]b[/sup]...)
Beyond that, though, no idea how to help on that problem. Sorry. It's been two years since I did chemistry in any way, shape, or form. Hope it helps, though.[footnote]Wikipedia, Equlibrium Constant [http://en.wikipedia.org/wiki/Equilibrium_constant][/footnote]
EDIT: Although, I'm thinking it will be the inverse. If the equilibrium constant of the first expression is K[sub]1[/sub], then the equilibrium constant of the second, K[sub]2[/sub], should be
K[sub]2[/sub] = (K[sub]1[/sub])[sup]-1[/sup]
by virtue of the products and reactants having switched places in the expression.
EDIT2: Added my sources.
