Question

Nicole measured the pH of a 0.100 M solution of propanoic acid, CH3CH2COOH, a weak organic acid at equilibrium and found it to be 2.931 at 25C. Calculate the Ka of propanoic acid. Her lab instructor mentions that the half equivalence method is better for determining pKa. What is the half-equivalence point and why is this method better at determining pKa? What is the pH after 20.0 mL of 0.0750 M NaOH is added to 30.0 mL of the 0.100 M propanoic acid? Kw = 1.00 x 10^-14

Answer 1

Ht ANSFWR ell elty Coom Ž Ciaz Cty Coot - CHAD name (A-) Intially, AH 2.931 -log [N+) = 2931 One Cht) = 102.931 O Con lau" we know, THAT ſkal. So, 102.931 = ✓ Kat Onl 0% Koi 2:1.37 *105 (Ausules), the 6 Half equivalent means - Concentration of Acid = Concentration of salt So [HAJ = EAJ Handoroom equation — pole prend lang empat om Pl_pka Flags on (pH pk ] Bc. In this method ... pką of the Solution directly determine from the port of the solution, so This method is very easy and usefulla ③ Zomeo. IM HAŤ some 0.075M NOOR Buffer Solution form, CHA) = 30001 - 36 * 0075 Leden [Nata-J > 20 x0 (1, 075 50

It in Pointer t Na A = [HAT the Half en equiralance pu z pra PM-log (1.87X105) I Pa 24.86 answer added pH & 21.86 when goml 0.075 m Nash in to zome of Orth Propansie avid. 1 Please like and comment.