Answer:
(a) [tex]0.88\%[/tex]
(b) [tex]24.4\%[/tex]
Explanation:
Hello!
In this case, since the ionization of benzoic acid produces benzoate and hydrogen ions:
[tex]phe-COOH\rightleftharpoons phe-COO^-+H^+[/tex]
Whose equilibrium expression is:
[tex]Ka=\frac{[phe-COO^-][H^+]}{[phe-COOH]}[/tex]
Whereas Ka is its acid ionization constant which equals 6.25x10⁻⁵, and each concentration can be expressed in terms of the reaction extent ([tex]x[/tex]):
[tex]6.25x10^{-5}=\frac{x^2}{[phe-COOH]_0-x}[/tex]
In such a way, for the two given initial concentrations, 0.80 M and 0.00080 M we compute [tex]x[/tex]:
(a) 0.80 M
[tex]6.25x10^{-5}=\frac{x^2}{0.80-x}\\\\x=0.00704M[/tex]
(b) 0.00080 M
[tex]6.25x10^{-5}=\frac{x^2}{0.00080-x}\\\\x=0.000195M[/tex]
Thus, the percent ionization for each case turns out:
(a)
[tex]\% ionization=\frac{0.00704M}{0.80M} =0.88\%[/tex]
(b)
[tex]\% ionization=\frac{0.000195}{0.00080M} =24.4\%[/tex]
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