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40% G Thu 4:24 PM a Chrome File Edit View History Bookmarks People Window Help T Telemundo Watch Full Episod x Lehman College CHE 168 S x Lehman College CHE 168 S x www.saplinglearning.com biscms/modibis/view.php?id 3352106 Apps All concerns D Dashboard l Gene M WileyPLUS Sapling Learning a Valentina L.A. Wom. a The NuBra Seamless Jumeirah Inside 3... WP Online Eyeglasses. B Other Bookmarks a Attempts Score 2/26/2017 11:55 PM 8.3/10 2/10/2017 01:12 PM Gradebook Assignment Information Print calculator Periodic Table Available From Not Set 00 Question 6 of 6 2/26/2017 11:55 PM Due Date 00 Map General Chemistry th Edton Points Possible City University of New York ICUhm, Lehm 00 Grade Category: Topic 5 Home Total (10 Phosphoric acid, H3PO4(ag), is a triprotic acid, meaning that one pts molecule of the acid has three acidic protons Estimate the 00 the concentrations of all species in a 0.100 M phosphoric acid solution, 6,9x103 6,2x108 4,8x10 Description 00 Number Policies Practi H,PO Number You can check your answers. You can view solutions when you complete or give up on any question. Number You can keep trying to answer each question until OH you get it right or give up Number There is no penalty for incorrect answers. HPOT Number O pH eTextbook O Help With This Topic OWeb Help & Videos O Technical Support and Bug Reports Scans-02-15.2017-1-epor Scans-02-15.2017-1-pdf Show

Explanation / Answer

H3PO4     ---------------->   H2PO4- + H+

0.10 0          0

0.10 - x                                    x           x

Ka1 = [x][x]/[0.1-x]

6.9 x 10^-3 = x^2/(0.10-x)

x^2 + 6.9 x 10^-3 x - 6.9 x 10^-4 = 0

x = 0.023

[H3PO4] = 0.10 - 0.023 = 0.077 M

[H3PO4] = 0.077 M

[H2PO4-] = x

[H2PO4-] = 0.023 M

[H+] = x = 0.023 M

almost maximum H+ ions comes from first ionisation so

pH = -log [H+] = -log(0.023)

pH= 1.64

[OH-] = kw/[H+] = 1 x10^-14/0.023

[OH-] = 4.35 x10^-13 M

second ionisation constant value always =[HPO4^-2]

[HPO4^-2] =6.2 x10^-8 M

  HPO4^2 ---------------------> PO4^-3 + H+

6.2 x10^-8                            0             0.023

6.2 x10^-8- z                        z                0.023 +z

Ka3 = [ PO4^-3][H+]/[HPO4^2]

4.8 x 10^-13 = (z)x(0.023 +z)/ (6.2 x10^-8- z )

z= 1.29x 10^-18 M

[PO4^-3] = z= 1.29 x 10^-18 M

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