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liz.goetzke liz.goetzke
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Posts: 6
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12 years ago
I would appreciate your unique working style to resolve this problem.
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The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO) is 3.0 × 10-8. At equilibrium, the molarity of H3O+ in a 0.010 M solution of HClO is __________.

a. 1.7 × 10-5    
b. 0.010    
c. 5.8 × 10-10    
d. 4.76      
e. 2.00
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wrote...
#1
12 years ago
HClO <>H+ + ClO-
initial concentration
0.010.......0.......0
at equilibrium
0.010-x.....x.......x

K = 3.0 10^-8 = (x)(x)/0.010-x

x=0.0000173 M

pH = -log 0.0000173= 4.76
wrote...
#2
12 years ago
1) The reaction of dissociation for hypochlorous acid is:

HClO + H2O Leftwards Arrow------> H3O+  + ClO-

2) The acid-dissociation constant is then defined as follows:

Ka = [H3O+][ClO-] / [HClO]

we know that:
Ka = 3.0 x 10-8
[H3O+] = x
[HClO] = 0.010 M
[ClO-] = x (according to the reaction above acid disociates equimolarly in both species: the hydronium and the hypochlorite ion)

3) We can substitute:

3.0 x 10-8 = x² / 0.010M

we solve for x:

x = 1.73 x 10-5 M

That is the H3O+ concentration at equilibrium (option a) !!!

Good luck!
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