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Chrome File Edit View History Bookmarks People Window Help University of Minnesota. D x-eChegg Study | Guided Sol X c Dwww.saplinglearning.com/ibiscms/mod/ibis/view.php?id=2200617 My Assignment Resources 10/27/2015 1 1:55 PM 0 0/10 Assignment Information Available From 10/26/2015 02:35 Due Date: Points Possible: Grade Category: Graded Description: Policies: Gradebook Attempts Score Print Calculator Periadic Table Question 1 of 3 10/27/2015 11:55 Map 10 sapling learning For the diprotic weak acid H2A, Kal-3.0 x 10%, and Ka2-5.0 × 10-9 What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A in this solution? Homework You can check your answers. You can view solutions when you com up on any question. You can keep trying to answer each q you get it right or give up. You lose 5% of the points available to in your question for each incorrect att answer Number Help With This Topic Web Help & Videos Technical Support and Bug Reports @ Previous Give Up & View Solution O Check AnswerNext Exit Hint Copyright © 2011-2015 Sapling Learning, Inc.-158 about uscareersprtnesprivacy policy terms of usecontact us help

Explanation / Answer

H2A <==> HA- + H+

let x amount has dissociated,

Ka1 = 3 x 10^-6 = x^2/(0.07-x)

x = [H+] = 4.58 x 10^-4 M

pH = -log[H+] = 3.34

[H2A] = 0.07 - 4.58 x 10^-4 = 0.0695 M

HA- <==> A2- + H+

let x amount has dissciated

Ka2 = 5 x 10^-9 = x^2/4.58 x 10^-4

x = [A^2-] = 1.51 x 10^-6 M

So we have at equilibrium

pH = 3.34

[H2A] = 0.0695 M

[A^2-] = 1.51 x 10^-6 M

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