Respuesta :
Answer:
pH = 1.39
Explanation:
Given that:
The molarity of [tex]HClO_2[/tex] = 0.15 M
with acid dissociation constant [tex]K_a[/tex] = [tex]1.1 \times 10^{-2}[/tex]
Acid dissociation constant of HClO = [tex]2.9 \times 10^{-8}[/tex]
Molarity of HClO = 0.15 M
The objective is to determine the pH of the solution:\
To determine the concentration of [tex]H_3O^+[/tex] obtained from both acids
The equation for the reaction can be expressed as :
[tex]HClO_2 + H_2O \to H_3O^+ + ClO_2^-[/tex]
The dissociation constant for the above reaction is as follows:
[tex]ka = \dfrac{ [H_3O^+] [ClO_2^-]} { [HClO_2]}[/tex]
[tex]1.1 \times 10^{-2} = \dfrac{ [x] [x]} { [0.15]}[/tex]
[tex]1.1 \times 10^{-2} = \dfrac{ [x]^2 } { [0.15]}[/tex]
[tex]x^2 = 0.15 \times 1.1 \times 10^{-2}[/tex]
[tex]x^2 = 0.00165[/tex]
[tex]x=\sqrt{ 0.00165}[/tex]
x = 0.04062 M
Now to determine the concentration of [tex]H3O^+[/tex] obtained from HClO
The equation for the reaction can be expressed as :
[tex]HClO + H_2O \to H_3O^+ + ClO^-[/tex]
[tex]ka = \dfrac{[H3O^+] [ClO^-]}{ [HClO]}[/tex]
[tex]2.9 \times 10 ^{-8} = \dfrac{[x] [x]}{ [0.15]}[/tex]
[tex]2.9 \times 10 ^{-8} \times [0.15] = {[x]^2}[/tex]
[tex]{[x]} ^2=4.35 \times 10^{-9}[/tex]
[tex]x=\sqrt{4.35 \times 10^{-9}}[/tex]
[tex]x=6.595 \times 10^{-5}[/tex]
Thus the total concentration now is :
x = [tex]H_3O^+[/tex] = [tex]0.04062 + 6.595 \times 10^{-5} M[/tex]
[tex]H_3O^+[/tex] = 0.04068595 M
[tex]H_3O^+[/tex] [tex]\simeq[/tex] 0.04069 M
pH = -log [H3O⁺]
pH = -log [0.04069]
pH = 1.39
The pH of a solution is 1.39. The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.
Given:
The molarity of HClO₂ = 0.15 M
Acid dissociation constant of HClO₂ = [tex]1.1 *10^{-2}[/tex]
Acid dissociation constant of HClO = 2.9 × 10-8
Molarity of HClO = 0.15 M
Determination of pH:
The equation for dissociation of HClO₂ can be given as:
[tex]HClO_2+H_2O--- > H_3O^++ClO_2^-[/tex]
The dissociation constant for the above reaction is as follows:
[tex]Ka = \frac{[H_3O^+][ClO_2^-}{[HClO_2]} \\\\1.1*10^{-2} =\frac{x*x}{0.15}\\\\1.1*10^{-2} =\frac{x^2}{0.15}\\\\x^2=1.1*10^{-2}*0.15\\\\x^2=0.00165\\\\x=0.04062 M\\\\[/tex]
Now to determine the concentration of [tex]H_3O^+[/tex] obtained from HClO.
The equation for the reaction can be expressed as :
[tex]HClO+H_2O--- > H_3O^++ClO^-[/tex]
[tex]Ka = \frac{[H_3O^+][ClO^-]}{[HClO]} \\\\2.9*10^{-8} =\frac{x*x}{0.15}\\\\2.9*10^{-8} =\frac{x^2}{0.15}\\\\x^2=2.9*10^{-8} *0.15\\\\x^2=4.35*10^9\\\\x=6.595*10^{-5} M\\\\[/tex]
Thus, the total concentration now is 0.04069 M
Calculation for pH:
pH = -log [H3O⁺]
pH = -log [0.04069]
pH = 1.39
Find more information about Equilibrium constant here:
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