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c) A student titrated 50.0mL of a 0.10M solution of a certain weak acid with NaOH(aq) . The results are given in the graph above. (i) What is the approximate pKa of the acid?

Respuesta :

Answer:

3.2

Explanation:

The missing graph is attached in the image below:

Initially, p.H from the graph is approximately 2.1

Therefore; the hydrogen ion concentration is

[tex][H^+] = 10^{-2.1}[/tex]

[tex][H^+] = ({k_a \times 0.10})^{\dfrac{1}{2}}[/tex]

[tex]{k_a} = 6.3 \times 10^{ -4}[/tex]

pKa = [tex]-log ( 6.3 \times 10^{-4})[/tex]

pKa = 3.2

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