Question

The dissociation of acetic acid, CH3COOH, has an equilibrium constant at 25C of 1.8 x 10-5. The reaction is as follows : CH3COOH (aq)  CH3COO- (aq) + H+ (aq) If the equilibrium concentration of CH3COOH is 0.46 moles in 0.500 L of water and that of CH3COO- is 8.1 x 10-3 moles in the same 0.500 L, calculate [H+] for the reaction. 

Answer 1

1.8 × 10^-5 = $\frac {(8.1 × 10^-3 moles/0.500L)(H^+}{0.46 moles/0.500L}$$H^+ = 1.0 × 10^{-3}M$