A buffer was made with HC2H3O2 and its conjugate base. What is the pH of the buffer if at equilibrium there was[HC2H3O2] = 0.5000 M and [C2H302ª] = 0.7000 M and the Ka = 1.60 x 10-5 ?everything is based upon the acid dissolving in water:HC2H3O2+ H20 <-> C2H3O21 + H30*1[Ka =[solve for [H30*'] then put in the numbersKa [([H3O+1] :[][H3O+'] =pH = - logd. HC6H6O61f. SO3?a. H30+1b. Он1c. H2C6H606е. HSO31-2g. H2SO3h. HC2H3O2i. C2H3O21j. H2Sk. HS-1I. s2m. HXп. X-1О. 1р. 2q. 3r. 0.5000s. 5.5 X 10-5t. 3.5 X 10-4u. 8.00 x 10-5v. 1.75 x 10-6w. 1.11 x 10-8х. 5.5 х 102y. 0.7000z. 1.60 x 10-5аа. 1.143 х 10-5bb. 0.4900сс. 1.104 х 10-5dd. 0.7100ее. 1.26 х 10-5 ff. 4.901gg. 7.93hh. 4.942ii. 4.957

According to Henderson equation pH= pKa(HC2H3O2)+ log[C2H3O2-1]/ [HC2H3O2]

A buffer was made with HC2H3O2 and its conjugate base. What is the pH of the buffer if at equilibrium there was [HC2H3O2] = 0.5000 M and [C2H3021] = 0.7000 M and the Ka = 1.60 x 10-5 ? everything is based upon the acid dissolving in water: HC2H3O2 + H20 <-> C2H3O21 + H3o+1 [ Ka = solve for [H30*1] then put in the numbers Ka [ [H3O*1] =- [ [H30*1] = pH = - log

A buffer was made with HC2H3O2 and its conjugate base. What is the pH of the buffer if at equilibrium there was [HC2H3O2] = 0.5000 M and [C2H3021] = 0.7000 M and the Ka = 1.60 x 10-5 ? everything is based upon the acid dissolving in water: HC2H3O2 + H20 <-> C2H3O21 + H3o+1 [ Ka = solve for [H301] then put in the numbers Ka [ [H3O1] =- [ [H30*1] = pH = - log

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter14: Principles Of Neutralization Titrations

Section: Chapter Questions

Problem 14.32QAP

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