Question

for a 1.0×10−4 m solution of hclo(aq), arrange the species by their relative molar amounts in solution.

Answer 1

The order of relative molar amounts in solution is; H2O >>> HClO > H3O+ = ClO- > OH-

The equation is;  HClO  + H2O   ⇄   H3O+ +  OCl-

Now we must set up an ICE table as shown below;

   HClO       + H2O   ⇄   H3O+ +  OCl-

I     1.0x10^-4                          0             0

C        -x                                 +x           +x  

E  (1.0x10^-4 - x)                    x             x

Ka = [x] [x]/ (1.0x10^-4 - x)  

But Ka for HClO4 = 2.9 x 10^-8

Hence; x = 1.688 x 10^-6 M

So; x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77

But pOH = 14 - pH = 14 - 5.77 = 8.23

pOH= -log[OH-]

[OH-] = Antilog( -8.23)

[OH-] = 5.89 × 10^-9 M

For  [HClO4] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x 10^-5 M

And [H3O+] = [ClO-] = x =1.688 x 10^-6 M

Therefore; H2O >>> HClO > H3O+ = ClO- > OH- (water is present in large excess)

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In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O