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Chrome File Edit View History Bookmarks People Window Help D D 4) 80% (a, Sun 5:33 AM Q EE http C https:// black board.olemiss.edu /bbcswebdav/pid-1503963-dt-content-rid-38733638 1/courses/Chem 106 Web 1 DAVIS 2014-2015 SUM2/cp4.pdf E App untitled folder € App d If Facebook Twitter W Wikipedia N Yahoo! News Popular Imported From Safari Challenge Problem #4 Consider a diprotic acid, H2A, with the following Ka values. Kal 0.01 Ka2 0.008 If you have a 0.01 M solution of H2A, what is [H3O+] and the pH Hint: The Ka values are too close together to ignore the second equilibrium

Explanation / Answer

H2A -----------------------> H+   + HA-

0.01                                0           0--------------------initial

0.01-x                           x              x ----------------equilibrium

Ka1 = [H+][HA-] / [H2A]

0.01 = x^2 / (0.01-x)

x^2 + 0.01 x - 1 x 10^-4 = 0

x = 6.2 x 10^-3

x = [H+] = [HA-] = 6.2 x 10^-3 M

HA- -------------------------> H+   +         A-2

6.2 x 10^-3                      6.2 x 10^-3     0 --------------------> initila

6.2 x 10^-3-y                  6.2 x 10^-3+y    y --------------------> equilibrium

Ka2 = (6.2 x 10^-3+y )(y)/(6.2 x 10^-3-y)

0.008 = (6.2 x 10^-3+y )(y)/(6.2 x 10^-3-y)

4.96 x 10^-5 -0.008 y = 6.2 x 10^-3 y +y^2

y^2 + 0.0142 y - 4.96 x 10^-5 = 0

y = 2.9 x 10^-3

[H3O+] = y = 2.9 x 10^-3

[H3O+] = 2.9 x 10^-3 M

pH = -log [H3O+]

pH = -log ( 2.9 x 10^-3 )

pH = 2.54

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