m a 0.0100 M sulfuric acid (kasoa) solution. becomes more significant because th

Question

m a 0.0100 M sulfuric acid (kasoa) solution. becomes more significant because the percent o 0100 M, you can't neglect the Ho HSO, contributions from contribution results both EP is from the first centration To zation step is 0.0100 M ICE table ine the of for the second step in onization be 0100 M (due to the 0.0100 reaction) nitial ons for the equilibrium 0.0100 0.0100 step concentrations (from e x is small approximation SO2 12) is not the and to the initial (0.0100 x)x multiply out the expression 0.0100 to arrive 0.012 0.0100x 0.0100 x 0.012(0.0100 x) 0.0100x x2 0.00012 0,012x 0.0100x x2 0.022x 0.00012 0 sing the quadratic formula. 2a (0.022) t v (0.022)2 4(1) (-0.00012) 2(1) -0.022 t 0.031 0.027 or x 0.0045 Since x represents a concentration, and since concentration be negative, we reject the negative root. x 0.0045 m the calculated value of x [H3O+] 0.0100 x cond step produces almost 0.0100 0.0045 amount that must not be th dilute H2SO4 solutions. 0.0145 M -Continued on the net

Explanation / Answer

Fidn pH and [SO4-2] for a 0.0075 M of sulfuric acid

so:

H2SO4 is a strong acid so:

H2SO4 <--> H+ + HSO4- KA = infinite

HSO4- <-> H+ + SO4-2 Ka2 = 0.012

so..

[H+] initial = 0.0075

[HSO4-] = 0.0075

[H2SO4] = 0.0075 - 0.0075 = 0 (all is in HSO4- state)

so..

in equilibrium

[H+] = 0.0075 + x

[SO4-2] = x

[HSO4-] = 0.0075 - x

substitute in KA2

0.012 =(0.0075 + x)(x) / (0.0075 - x)

solve for x

0.012 *0.0075 - 0.012 x = 0.0075x + x ^2

x^2 + (0.0075 +0.012 )x - 0.012 *0.0075 = 0

x^2 + 0.0195x - 0.00009 = 0

x = 0.00385

pH = -log(0.00385) = 2.41

so

[SO4-2] = x = 0.00385 M

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