What is the ph of a sodium acetate (nac2h3o2) solution prepared by adding 0.820 grams of sodium acetate to 100.0 ml of water at 25.0 °c? the ka at 25.0 °c for acetic acid is 1.8 ⋅ 10-5?
According to the reaction equation: CH3COONa+ H2O ↔ CH3COOH + OH- when we have 0.82 g of sodium acetate in 100 mL So we have 8.2 g per liter & when we have the molar mass of CH3COONa=82 g/mol we have to get the molarity of CH3COONa = weight/molar mass = 8.2 / 82 =0.1 M So CH3COONa + H2O ↔ CH3COOH + OH- initial c 0.1 0 0 equilibrium C (X-0.1) X X when Kb= Kw / Ka and we have Kw = 1x10^-14 & Ka = 1.8x10^-5 So Kw/Ka = [CH3COOH][OH-] / [ CH3COONa] (1x10^-14)/(1.8x10^-5) = X^2 / (X-0.1) 5.6x10^-10 = X^2/(0.1-X) 5.6x10^-11 - 5.6x10^-10 X = X^2 ∴X= 7.48x 10 ^-6 ∴[OH] = 7.48x10^-6 M when POH = -㏒[OH] = -㏒(7.48x10^-6) = 5.13 ∴PH = 14 - POH = 14 - 5.13 = 8.87
