Answer:
[tex]pH=4.6[/tex]
Explanation:
Hello,
In this case, the dissociation of hypochlorous acid is:
[tex]HClO_4\rightleftharpoons H^++ClO_4[/tex]
Therefore, the law of mass action for equilibrium is written as:
[tex]Ka=\frac{[H^+][ClO_4^-]}{HClO_4} =3.00x10^{-8}[/tex]
And can also be written introducing the reaction extent ([tex]x[/tex]):
[tex]3.00x10^{-8}=\frac{x*x}{0.0200-x}[/tex]
Thus, solving for [tex]x[/tex] we obtain:
[tex]x=2.45x10^{-5}M[/tex]
Which is also equal to the concentration of H⁺ in the solution. Thereby, the pH turns out:
[tex]pH=-log([H^+])=-log(2.45x10^{-5})\\\\pH=4.6[/tex]
Which is substantiated by the fact it is about an acid (pH lower than 7).
Regards.
pH=4.6
Given:
Ka= [tex]3.00*10^{-8}[/tex]
[HClO]=0.0200 M
The dissociation of hypochlorous acid is:
[tex]HOCl[/tex] ⇌ [tex]ClO^- +H^+[/tex]
Therefore, the law of mass action for equilibrium is written as:
[tex]K_a= \frac{[H^+] [ClO^-]} {[HOCl]} \\[/tex]
At equilibrium [tex][H^+] \text{and} [ClO^-][/tex]be x and [HClO] be [0.0200-x]
On substituting the values in given equation we will get:
[tex]3.00 * 10^{-8}=\frac{[x]*[x]}{[0.0200-x]}\\\\3.00 * 10^{-8}=\frac{x^2}{[0.0200-x]}\\\\x=2.45*10^{-5} M[/tex]
This is also equal to the concentration of H⁺ in the solution. Therefore, the pH turns out:
[tex]pH= -log[H^+] \\\\pH= -log [2.45 * 10^{-5}]\\\\pH=4.6[/tex]
Hence, the pH of an aqueous solution is 4.6.
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