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Alex787 [66]
3 years ago
12

What is the buffer component ratio, (bro-)/(hbro) of a bromate buffer that has a ph of 8.08. ka of hbro is 2.3 x 10-9?

Chemistry
1 answer:
Alla [95]3 years ago
3 0

pKa= -log(ka)

= -log(2.3*10^-9)

= 8.64

Now pH can be calculated using equation:

pH=pka+log(BrO-)/(HBrO)

8.08 =8.64+log(BrO-)/(HBrO)

log(BrO-)/(HBrO)=8.08-8.64

= -0.56

(BrO-)/(HBrO)= 10^-0.56

=0.275

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Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

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Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

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Thus, at each temperature we obtain:

K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

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