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Scilla [17]
3 years ago
12

Explain why unboiled water can be used to prepare an acetic acid solution, but not a naoh solution?

Chemistry
1 answer:
shtirl [24]3 years ago
8 0
NaOH is a hygroscopic compound, which means that it has a very strong affinity for water. Moreover, NaOH is also a stronger base than acetic acid is a strong acid. These two properties of NaOH make it necessary for different precautions to be taken while preparing a solution. 
On the other hand, acetic acid is not very reactive and may easily be prepared in a solution of unboiled water.
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Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2
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Answer:

Equilibrium constant of the given reaction is 1.3\times 10^{-29}

Explanation:

NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr....(K_{p})_{1}=5.3

2NO\rightleftharpoons N_{2}+O_{2}....(K_{p})_{2}=2.1\times 10^{30}

The given reaction can be written as summation of the following reaction-

2NO+Br_{2}\rightleftharpoons 2NOBr

N_{2}+O_{2}\rightleftharpoons 2NO

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N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr

Equilibrium constant of this reaction is given as-

\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}

=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})

=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}

=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}

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