Question

25. + -10.1 points 0/4 Submissions Used (a) Calculate the percent ionization of 0.00410 M propionic acid (Ka = 1.3e-05). % ionization = (b) Calculate the percent ionization of 0.00410 M propionic acid in a solution containing 0.0500 M sodium propionate. % ionization =

Answer 1

ca the mi ionization - propionic acid a CH₃ CH₂ cooh (a weak acids ionization = conc" of Htion initial conch of acid X100 Now given that 0.00410 M propionic acid q Ka = 1:3x105 + + Now we know 3 Ka . -> 1.3 X10 x2 cgcloH CH₃ CH₂COOH + H₂O H₂O4 + CH ₂ CH ₂ Coo initial of 0.0041 middil t = x 1. tx tx Final → 0.0041-X . tX +X [H30+] [CH3CH2C00 - ] Oooh [CH3 CH2C004] (0'0091-) here we eas can see that ka in very small , so we cansider (0.0091 -~) 2D as 0.00918 so, i-3x105- 0.0041 + x = 2:3x10° Putting all the value on equn_1 % ionization = 2.3x10 3 0.0041 = 5.6 L ii . ionization = 5.6%) -oa X 100

6) 241 M % ionization conch of H+ inidial conal of aeid. -51 X 100 sodium propionate = 0.05 M propionic acid = 0.0041 M Ka [propionic acid] =1.3 x 10° - Na CH₂ CH₂ coo 0.05 M Nat+ CH₂ CH₂ coo 0.05M 0.05 M and, H307 +(4,CH2007 CH3CH2COOH + H20 2 initial Medium final 0.0041 -x 0.0041-8 + x tx + x tx so, ma [H307) [CH3 CH2600] X(065+90-057x) [CH3CH2COOH] (0.0041-x) 0.05 + x x 0.5 We can see that Ka value in very small, so we consider (0.0041-x) agas 0.0041 . Bx105 0.05 x 0.0091 => x = 1.06x1ö G y conc" of Htion y. ionization = .06 x 100_ - X100 y 0.0091 = 0.026 Y :.