Respuesta :
Answer: pH of a 0.500 M [tex]HNO_2[/tex] solution is 1.82
Explanation:
[tex]HNO_2\rightarrow H^++NO_2^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.500 M and [tex]\alpha[/tex] = ?
[tex]K_a=4.6\times 10^{-4}[/tex]
Putting in the values we get:
[tex]4.6\times 10^{-4}=\frac{(0.500\times \alpha)^2}{(0.500-0.500\times \alpha)}[/tex]
[tex](\alpha)=0.030[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex][H^+]=0.500\times 0.030=0.015[/tex]
Also [tex]pH=-log[H^+][/tex]
[tex]pH=-log[0.015]=1.82[/tex]
Thus pH of a 0.500 M [tex]HNO_2[/tex] solution is 1.82
The pH of a 0.500 M solution is 1.82
Balanced chemical equation:
[tex]HNO_2---- > H^++NO_2^-[/tex]
Initial: cM 0 0
Equilibrium: [tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
Dissociation constant:
[tex]K_a=\frac{(c\alpha)^2}{c-c\alpha}[/tex]
Given:
c= 0.500 M and [tex]\alpha[/tex] = ?
[tex]K_a=4.6*10^{-4}[/tex]
On substituting the values:
[tex]K_a=\frac{(c\alpha)^2}{c-c\alpha}\\\\4.6*10^{-4}=\frac{(0.500\alpha)^2}{0.500-0.500\alpha}\\\\\alpha=0.030[/tex]
[tex]H^+=c*\alpha\\\\H^+=0.500*0.030\\\\H^+=0.015[/tex]
Calculation for pH:
[tex]pH=-log [H^+]\\\\pH= -log [0.015]=1.82[/tex]
Thus, the pH of a 0.500 M solution is 1.82.
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