4. (20pts) Consider a mixture of 50.0 mL of 0.100 M HCl and 50.0 mL of...

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

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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