0.1M OF NaOH in 5ml is mixed with 0.1M of CH3COOH in 45ml, what is the concentration of CH3COO- in equilibrium mixture?

Answer 1

mols CH3COOH = 0.1 x 0.045L = 0.00450
mols NaOH = 0.1 x 0.005 = 0.0005

.....NaOH + CH3COOH ==> CH3COONa + H2O
I..0.0005....0.0045.......0..........0
C.-0.0005...-0.0005....-0.0005...+0.0005
E....0.......0.0040.....0.0005....0.0005

(CH3COO^-) = mols/L = 0.0005/0.050 = ?

Answer 2

thanks DRBob....how can i find the amount of CH3COOH undissociated in the equilibrium mixture?