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A 1.44 L buffer solution consists of 0.137 M butanoic acid and 0.275 M sodium butanoate. Calculate the pH of the solution following the addition of 0.069 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

Respuesta :

Answer:

The answer is "[tex]P^{H}=5.379[/tex]".

Explanation:

[tex]\ NaOH \ value = \frac{n}{v}\\\\[/tex]

                     [tex]=\frac{0.069\ moles}{0.144L}\\\\=0.04791\ M[/tex]

[tex]\ Ka=1.52 \times 10^{-5}\\\\P^{ka} = -10g \ ka \\\\[/tex]

      [tex]= -10 \times 1.52 \times 10^{-5}\\\\= 4.82\\[/tex]

Equation:

[tex]CH_3CH_2CH2COOH+NaOH\rightarrow CH_3CH_2CH_2COONa +H_2O\\\\[/tex]

[tex]\boxed{\left\begin{array}{ccccc}I &0.137 &0.04791 &0.275 & -- \\ C &-0.04791 &-0.04791 &+0.04791 & -- \\E &0.08909 &0&0.32291 & -- \end{array}\right}[/tex]

[tex]P^{H}= P^{ka}+\log\frac{CH_3CH_2CH_2COONa}{CH_3CH_2CH_2COOH}\\\\[/tex]

      [tex]= 4.82+\log\frac{0.32291}{0.08909}\\\\=5.379[/tex]

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