Question

4. (20pts) Consider a mixture of 50.0 mL of 0.100 M HCl and 50.0 mL of 0.100 M acetic acid. Acetic acid has a K. of 1.8 x 10-5. a. Calculate the pH of both solutions before mixing. b. Construct an ICE table representative of this mixture. c. Determine the approximate pH of the solution. d. Determine the percent ionization of the acetic acid in this mixture.

Answer 1

pH = solution (a) pH of the tond 0.1 m Hal we know, -log [H1] As Hel is completely dissociate so [HI -log[..] pH = 1.00 0.1 M PH W pit of the 50 ml 0. 100 M Cl Coolt given ka = 118810-5 AS CH coolt is weak acid, the pH formula for ch coolt is. Ipka -Ź loge Ź [-logka] - 1 log C P+ = {E-log (18210-5)]*] (0.4) 2.372 +0.5 2.872 PH. Som pH of AcoH = 2.872 Scanned with CamScanner

50 ml 0.100m Hel + 50 ml 0.1004 Ch Coolt, pH of the mixture is determina by NIVI+N₂ V2 = WRER here, WI, W2, ARE Concen. - truticg gf zothacia and resultant mixture ۷۱/ 72/4/3 - volume of both acid aci, and resultan mintire already know that, pit or o.lm Hel plt of O.IM Cl} co014 = 2.872 EDI Wi=Chal [H] = 10-1 W₂ = calcoolt via 50ml V2 = we = 1 - 1o-2.872 V2 = 50ml VR = 100ml 50 X 10-!. + 50 x 10-2872 = 100 X NR 5 + 0.0671 =NR 100 NR - @ 5.0671 100 = 0.0506 PH - log (0.0506) - 1.29 e scaned win PH of the mixture = 1.2g.

50 ml 0.100m Hel + 50 ml 0.1004 Ch Coolt, pH of the mixture is determina by NIVI+N₂ V2 = WRER here, WI, W2, ARE Concen. - truticg gf zothacia and resultant mixture ۷۱/ 72/4/3 - volume of both acid aci, and resultan mintire already know that, pit or o.lm Hel plt of O.IM Cl} co014 = 2.872 EDI Wi=Chal [H] = 10-1 W₂ = calcoolt via 50ml V2 = we = 1 - 1o-2.872 V2 = 50ml VR = 100ml 50 X 10-!. + 50 x 10-2872 = 100 X NR 5 + 0.0671 =NR 100 NR - @ 5.0671 100 = 0.0506 PH - log (0.0506) - 1.29 e scaned win PH of the mixture = 1.2g.

(d) Helt tal- this oil conres foon te comen -tration ion of Hel Che coolt chcooottt (0.1-x) Х X x(x+0.0 ka= (0.1–) hereka=1.8x10-5 X (x+0.1) 1.8 x 10-5 CO? neglecting other team 24x ot X = 1.8 X10-5 poof percent of ionization of Clb coolt in this mixture 1-8810-5 Xloo 0.1-x) = 1.8810-5 Gol 0.018 % Scanned with CamScanner