Question

Question 14

Calculate the pH of a 0.051M solution of sodium acetate (CH₃COONa). The Ka for lactic acid is 1.4 × 10^-4.

a. 12.71

b. 2.57

c. 1.29

d. 8.28

e. 11.43

Answer 1

dissolve completely in water. relation between ka of and acid and GS CH₃COONa CH₃COo + Not CH₃COO is a weak conjugate bosc or acetic acid CH₃COOH acetate in forms OH in reaction with water. CH₃COo + H₂O - CH3COOH + H₂O oh Now form an ICE table for the reaction and form found equilibrin concentration, um O - I DC DU CH₃COO + H2O = CH₃COOH toh Initial 0.051 change 0.051-C equilibrium Now, equilibrium constant for the reaction or bose dissociation constant of CH₃Coo is given by [CH3coot [ot] at equilibolur [CH₃Cool Kb² Kb of ets conjugate base is

Constant of water Kakb-kw kw is dissociation Кь 1014 - 1 7.143 x 10 ka Note a ko 1-4X154 Note . There is slight error in the question, Question should bove given Ko of CH₃COOH Cacetic acid) cohich is 1.8x10 s from tables. Other the salt should have been sodium lactate, which is conjugate base of lactic acid. But we proceeding with ka-14x154 for acetic acid because the answer is getting by this value) wise arc Now, Kb IC. 20 7. 143 x 10 (0.051-x) low than concentra- since kb is very tion of base, we can assum LLO.05 0.051-2 - O.osi. Or 2 0.051 = 7.14341" x Voos1x2.143 x16 3 1.91x106

Now, 6 Pota-log [on] - log (1.91810) X- COH) 1.918106 POH- 5.72 Now, PH+POH - 1 pH = 14-POH = 14-8.72 PH-8.28 2) a) 285 9 H3 Pou IS 400 ml Solution mass of solution. - volume x density 400 ml x 1.35 g ML 540g mass % of a solute is defined by mass % (or weight %) mass of solute X100 mass of solution 2859 Xivo 54og 52.778 % mass %= 52.8%

solution is defined 6) Polarity of a by molarity moles of solute Volume of solution in liter first we have to find moles of top mass moles molar ass. 2859 97.99519 mol 2.9083 mol Volume of solution=400 ml = 40oXIOL = 0.4 Liter. molarity 2.9083 moll 0.4 L = 4.27 mol Concentration in mol is 7.27 m