Question

6. (2 points) A student diluted 125 mL of 0.500 M acetic acid to a final volume of 500. mL and found the pH of this solution to be 2.82. Calculate the pk, and percent ionization of this acid. Show your calculations and give your answers to the correct number of sig figs.

Answer 1

Solution given data Vi = 125 ml - m = 0.500 m (Initial Come) - in Dilution - V V a V = 500ml m = 2 Final Cosic...). initial volume final volume mole of acetic acid after dilution) mole of acetic acid C. Before dilution & mv = move 0.5X125 = max 500 for mg = 0.125m. acetic acid, CH₂COOH - CH₂ COO + hoa {ag Cao tot coca ca ca 2 - Degree of Ionization ,

Page No. (kas equilibrium constant for this son Ra = [ch coo) TH Ген, сор) Ra = (2xca C (1-2) ka = ca 17a 1-2 = 1 Hint 2 - Ka 2. will be dat het very less for weak acid Htions concentration at equilibaium , [ht = ca. [nt] = Trac pH = -log[ut] pH = -log (kache pH = I [-logka - loge

PRO No. pH = 1 [pta logo . [ pka_ - logika [log now layon Alagning pH = 2.82 (given) c = 0.125 m C calculated) 2.82 a log (0.125) q 2:32 = b [pka - dog425) = -0.900 :) = - 0.900 pka t 0.90 4.74,

fencentage ionization (2) 2= | ka pka - - log ka = 4.74 ka = 1.88105 18x 105 0.125 2 = 12x10 3 = 0.012 Percentage ionization (7.2) = 0.012 X 100 = 1.2 % 1 % 2 = 1.2%