Question

0.099 al 1. Om 2.04 CA COOH Aldanty of CH COON cu Volume of CCORY Cm PAL First Beading po Second Reading) CHCOOH talle present Cmokes Concentration to equilibriunful H present @ Equilibrium moles) CHE COO Present equilibrium Conds) Concen. FCHCOO © Equilibrium cu Cance CHE COOH PROTECH Fresents @ Equilibrium (miles) Concen. CH3COOH @ Equilibrium cu 2.06 2.05 Aug DA colla
can you help me go through the steps of finding these questions up above ? the given concentration, volume of ch3cooh, and avg. ph are given. please show work and how units cancel. thank you

Answer 1

I am trying to solve your problem using values provided in question.
Solution- The given values are Melarity of CH3COOH Z 0-099 Malay volume of C43COOH = 10.0 ml PH z 2.05 Milimales of Chg CooH initially = 0.099X10 = 0.gg milimale Dette males of Chacool = 6-99X10-3 males for at equilibecium of 22:05 pH = -log [at] 2005 2 + log [+] -2.05 [n] z 10 Z- 102.05 28.9 concentration of Chr"] at equilibrium z 8.91 X163 m Ht present at equilibrium z 8.91x10 3 x 10 milimole z 89.1 x 15 males ** peussent at equilibrium 2–8.91xtos males Now equation of dissociation of orgcoon can be maritten os CH₂COOH - CH₂600 +H* 6 initial 0. 99 0 0 елшісі Вчи» 0-99 -1 2 x 171

milimales of Htat equilibrium z H= 8.91 X 162 milimole milionoles of cngcoó at equilibrium z az 891X15 Prilimole 28.91x105 mole Malls of c49c00 pecesent at equilibrium - 8:918105 male Concentration of chycoo at equilibseium = _8.91x162 millimole - Lomiltlitere = 8.91X153 Male ſeitser 28.91 X 10-3 Molar Milimales of Checcom at equilibrium = 0.99 – x 2099 - 0.0891 = 0.90 milimale 2o 9oX to 3 Male Moles of Che coon at equilibrium z 9.0 x1o4 mole Concentration of Cuzcook at equilibrium e 0.go mklimele 20 mililiter 2 9х10 “маа = 98102 Molar {from } Mitu Kaza (0.99-) TO the [CH₂600] [mt] 8.91X103 x 8.9 4X103 [CH₂COor 9x102 ka = 882x10-4 10.882x10-5