## If

The charge balance and proton condition, with H20 and HAc being the reference specjifc.v are identical: ... .'u-: . •. '■•;• ■ "•

I/--/:: ' ['};■ ■ ' ;-f. ':•":. [H+] = [OHI + [AC] .' '.7

y From the assumption noted above ([H+] [OH~]), the charge balance reduces to ' 'I-/ '

¿^r■ ' i'7 ■•. . • [H+J = [Ac-]/'i;'7 v."S

; If we let x = [H+] = (Ac"] and note from the mass balance on acetic acid that [HAc] = ? 0.01 - x,wp can substitute into the equilibrium relationship for acetic acid: : . ■

' Multiplying through and rearranging terms yields js2 + 1.8 X 10"sx — 1.8 X 10~7 = 0 V

This equation can be solved using the quadratic formula: i

-(1.8 X 1Q"5) ± Vq.8 x IP-5)2 - 4(1)(—1.8 X 10-Q

Thus, [H+] = 4.15 X 10~", or pH = 3.38. This is the same answer pbtained in'Example . ! 4 2 when simplifying assumptions were not used

;: It is clear that since pH =,3.38, our assumption that [H1] is much greater than [QHr] is a good one The second assumption that [HAc] is greater than [Ac-] is also a good one :'since [Ac-] = 4:14 X 10"4. From the mass balance on acetic acid: . ■■ "v:;;..^j*'J

; [HAc] = 0.01 - 4.14 X 10"4 = 9.59 X IP"3_

collections.' The strident should note that NaCl completely ionizes upon addition to water. For this problem, the equilibrium relationships are

I: The mass balance relationships are .';

Qhac = 0.01 = [HAc] + [Ac"] '; ■ , ' Cr.Na = 0.01 = [Na*] :. ."Vh. 'V: : .ct - °-01 = [en ■ v.": :-v'-

'[H+] + [Na+] = [OH"] + [Ac~] + [cr] ; . ■ . . : 