Question

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Start each problem by writing a balanced acid ionization in water reaction and setting up an appropriate ICE table. 1a 10.0 mL of 0.10 M acetic acid: pH = 2.87 K. = Step 1: Acid ionization reaction: HCH3CO2(aq) + H2O1) CH3CO2 (aq) + H2O*(aq) Step 2: Set up an ICE table HCH:CO2(aq) H2O CH3CO2 (aq) H3O*ac) - с E Step 3: Determine the [H3O*) using the pH. [H30%) = 10-PH Step 4: Fill in your known values in to the ICE table. Do CiV1 = C2V2 dilutions if necessary, to find concentrations. (Hint: this isn't necessary in question 1a, but is necessary if you're mixing your acid with anything else). Step 5: Write the equilibrium expression using your balanced equation from Step 1. Solve for Ka.

Answer 1

La) 10 ml of 0.1M acetic acid : pH = 2.87 Ka 1.75x10-5 step1 H CHg CO2Qq.) + H2O com] CH₂CO (aq) + 4.8 Step 2 | H CH ₂ CO 2 c09) H₂Oce) CH₃ CO ₂ cag) H3ot (aq) I 0.1 С -e 2 e E O. 1-x fe fr Step 3 -2.87 [Hot] = 10 1.35x10 M = 0.00135 M Step 4 add 10 ml of o.1M NaOH then CH₂COo Healthy > CH3COONa +H2O Od t Naor (og) CH3Cool NaoH cag.) CH₃Coona H2O I o. C u E 0.1-x 0.1-2 of a Step 5 CH₃COo + H + H+ снасоон at equilibrium (0.1->) 2 20 Ka - [CH3 Cool] [H+] [CH3COOH] e x x o.1-se 22 x2 Co.)-c) * 0.1 011-2 Ka= x2