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Chemistry
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a. Calculate the pH of a 7.71×10-3 M solution of H2SO4

(Ka = 0.0120 for HSO4-)

b. Calculate [OH-] of a 5.45×10-1 M aqueous solution of pyridine (C5H5N, Kb = 1.7×10-9).

c. Calculate the pH of the above solution.

Feb 13th, 2015

a. Since H2SO4 is a strong acid [H+]=[H2SO4], Ka for bisulfate is irrelevant. 

pH=-log(7.71*10^-3) = 2.113

b. We have the reaction

              Py         +  H2O   ->   OH-  +  PyH+    ; Kb=1.7*10^-9

i           0.545                            0            0

c            -x                               +x          +x

e         0.545-x                          x             x

Kb=[OH-][PyH+]/[Py]     =>    (1.7*10^-9)=(x)(x)/(0.545-x)

=>   (x^2) = (9.3*10^-10) - x(1.7*10^-9)   => (x^2) + (1.7*10^-9)x - (9.3*10^-10) = 0

Using the quadratic formula we get x = (3.1*10^-5) hence [OH-] = 3.1*10^-5 M

c. pOH= -log(3.1*10^-5) = 4.5

and pH + pOH = 14 => pH = 14 - pOH = 14 - 4.5 = 9.5 

Feb 13th, 2015

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Feb 13th, 2015
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Feb 13th, 2015
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