Answer
(a) $K_c = 1.38 \times 10^{-4}$
(b) $[C_3H_6O_3] = 0.992 M$
Work Step by Step
** Remember to put the balanced coefficients as the exponents of the concentrations.
(a)
$K_c = \frac{[Products]}{[Reactants]} = \frac{[H^+][C_3H_5O_3^-]}{[C_3H_6O_3]} = \frac{(3.65 \times 10^{-3})(3.65 \times 10^{-3})}{(9.64 \times 10^{-2})} = 1.38 \times 10^{-4}$
(b) Now that we know the equilibrium constant, substitute the values of the given concentrations, and solve for $[C_3H_6O_3]$:
$1.38 \times 10^{-4} = \frac{[H^+][C_3H_5O_3^-]}{[C_3H_6O_3]} = \frac{(1.17 \times 10^{-2})(1.17 \times 10^{-2})}{[C_3H_6O_3]}$
$[C_3H_6O_3] = \frac{(1.17 \times 10^{-2})(1.17 \times 10^{-2})}{1.38 \times 10^{-4}} = 0.992$
