Acetic Acid = 0.1033 MKa = 1.8 x 10-5Sodium hydroxide = 0.1104 MThe first picture are the data, if neededsecond picture are the ones needs to be solved HC2H3O2H++C2H302-Initial0.1033ChangeEquilibrium1. Initial concentration [HC2H3O2]o = 0.1033 M-X+X+X0.1033-xX[H*][C2H307]2. Ka symbolic expression: Ka[HC2H302]x23. Ka numerical equation with x: 1.8 x 10-50.1033-x4. Quadratic formula methoda. quadratic equation: x2 + 1.8x10-5x – 1.86x10-6 = 0%3D-1.8x10-5+V (1.8x10-5)2-4(1)(-1.86×10-6)b. quadratic formula: x =2(1)x = 1.35x10-3 M; -1.37x103 M (only positive root is used)5. Assumption (5% rule) methoda. equation for x: x =V(1.8 x 10-5)(0.1033) = 1.36 × 10-3M1.36x10-3 Mb. % ionization =x 100 = 1.32%0.1033 M6. Equilibrium concentrationsa. [HC2H3O2leg = 0.1033 – 1.35x10-3 = 0.1020 M%3D%3Db. [H*]eg = 1.35x10-3 MJeqc. [C2H3O2]eg = 1.35x10-3 M7. pH = -log[H*] = -log(1.35x10 3) = 2.87 B. Calculate the pH after 17.50 mL of N2OH is added to 35.00 mL of acetic acid using the concentrationassigned in Beyond Labz:• First assume the acid-base neutralization reaction is irreversible and goes to completion:田HC;H;O2 (ag)+ Он (ag)→ H2O ()+ C;H3O2 (aq)Moles before reactionChange in molesMoles remaining after reactionThen equilibrium is establishedHC;H;O2 (ag)2H* (ag)+ C;H;O2 (aq)IInitial Concentration (M)Change in Concentration (M)Equilibrium Concentration (M)EEquilibrium Concentration,Calculated (M)Show your work here or replicate this on a separate piece of paper. Number and label each calculation, includeunits, and show the math performed to get your answer. Report answers with correct units and significantfigures.

Given: Concentration of Acetic Acid = 0.1033 M Concentration of Sodium hydroxide = 0.1104 M Volume…

B. Calculate the pH after 17.50 mL of NAOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: • First assume the acid-base neutralization reaction is irreversible and goes to completion: HC2H;O2 (ag) + OH° (ag) - H2O (1) + C;H;O;" (ag) Moles before reaction Change in moles Moles remaining after reaction • Then equilibrium is established HC;H;O2 (ag) 2H* (ag) + C;H;O2 (ag) Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M) Show your work here or replicate this on a separate piece of paper. Number and label each calculation, include units, and show the math performed to get your answer. Report answers with correct units and significant figures.

B. Calculate the pH after 17.50 mL of NAOH is added to 35.00 mL of acetic acid using the concentration assigned in Beyond Labz: • First assume the acid-base neutralization reaction is irreversible and goes to completion: HC2H;O2 (ag) + OH° (ag) - H2O (1) + C;H;O;" (ag) Moles before reaction Change in moles Moles remaining after reaction • Then equilibrium is established HC;H;O2 (ag) 2H* (ag) + C;H;O2 (ag) Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) E Equilibrium Concentration, Calculated (M) Show your work here or replicate this on a separate piece of paper. Number and label each calculation, include units, and show the math performed to get your answer. Report answers with correct units and significant figures.

Chemical Principles in the Laboratory

11th Edition

ISBN:9781305264434

Author:Emil Slowinski, Wayne C. Wolsey, Robert Rossi

Publisher:Emil Slowinski, Wayne C. Wolsey, Robert Rossi

Chapter21: Rates Of Chemical Reactions, Ii. A Clock Reaction

Section: Chapter Questions

Problem 2ASA

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