(the answer is 8.94) sorry for all the questions! NaC2H3O2----- Na+ + C2H3O2- (rxn) this is a salt so you have to determine which piece to use for the next step, because if you add water to both pieces (Na + H2O= NaOH WE CANNOT MAKE STRONG) so you do not use this one you use the other one to start the problem. Visit BYJUS to study the uses, properties, structure of Acetate ( C2H3O2(−)) explained by … Problem: Ka for HC2H3O2: 1.8 x 10 -5Ka for HCO3-: 4.3 x 10 -7Using the Ka's for HC2H3O2 and HCO3, calculate the Kb's for the C2H3O2- and CO32- ions. The KB value is listed below: pKa + pKb - 14 pKa = -log10 (Ka), Ka of acetic acid = 1.8*10^-5 pKa = 4.74, pkb = 9.255 9.255 = -log10(Kb), Kb = 5.56*10^-10 Acetate (C2H3O2(−)) - Acetate is the chemical name of C2H3O2(−). Register to view this lesson. The Kb value is high, which indicates that CO_3^2- is a strong base. Kb=[C2H3O2H][OH-]/[C2H3O2-] *As always, water does not show up in the equilibrium equation. To find the Ka, solve for x by measuring out the equilibrium concentration of one of the products or reactants through laboratory techniques. Plugging in the values found for the equilibrium concentration as found on the ICE table for the equation Ka = [H3O+][C2H3O2]/[HC2H3O2] allows the value of Ka to be solved in terms of x. 3) Next, you need to find out what the terms you should use in the equilibrium expression, this is the most difficult step, so I'll break it into 2 parts a and b respectively. The equation is for the acid dissociation is HC2H3O2 + H2O <==> H3O+ + C2H3O2-. I've tried a bunch of times but I can't get the correct answers. [OH-] [C2H3O2-] (eq. Acid with values less than one are considered weak. Pt.) To determine the pH you will need the concentration of potassium acetate and the Ka (= 1.7e-5 for acetic acid). LiOH is a strong base, therefore it is 100% dissociated. An equilibrium constant requires that the base (or acid) is not 100% dissociated and that there is an actual equilibrium in solution. FREE Expert Solution Show answer 99% (168 ratings) It will have no Kb value because of that. Below Is The Data: Please Explain In Detail # Of Trials Initial V(HC2H3O2) Vol(NaOH) PH(equiv. Compare these Kb values to these: Kb for C2H3O2 = 4.38 x 10^ -12 Kb for CO3 = 1.99 x 10^ -5 Pt.) Strong acids are listed at the top left hand corner of the table and have Ka values >1 2. TABLE OF CONJUGATE ACID-BASE PAIRS Acid Base K a (25 oC) HClO 4 ClO 4 – H 2 SO 4 HSO 4 – HCl Cl– HNO 3 NO 3 – H 3 O + H 2 O H 2 CrO 4 HCrO 4 – 1.8 x 10–1 H 2 C 2 O 4 (oxalic acid) HC 2 O 4 – 5.90 x 10–2 [H 2 SO 3] = SO 2 (aq) + H2 O HSO Question: Calculate Kb The Base Dissociation Constant For [C2H3O2-], Acetate Anion, For Each Of Your Trials From The Concentrations Of Species At The Equivalence Point. The first one is: Determine Kb for a) the ethanoate ion, C2H3O2^- (aq) (the answer is 5.6 x 10^-10) b) the borate ion, H2BO3^- (aq) (the answer is 1.7 x 10^-5) The second one is: Calculate the pH of a 0.20 mol/L solution of a base, where Kb = 3.82 x 10^-10. this is potassium acetate, the salt of the weak acid ethanoic acid or acetic acid. 1.