Answer: pH of the solution is 2.6
Explanation:
[tex]HA\rightarrow H^+A^-[/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= 0.115 M and [tex]\alpha[/tex] = [tex]\frac{2.1}{100}=0.021[/tex]
[tex](\alpha)=0.030[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex][H^+]=0.115\times 0.021=0.0024[/tex]
Also [tex]pH=-log[H^+][/tex]
[tex]pH=-log[0.0024]=2.6[/tex]
Thus pH of the solution is 2.6
The pH value of the solution is : 2.6
Given data:
percentage deprotonation = 2.1%
concentration of solution ( c ) = 0.115 M
Determine the pH of the solution
Applying dissociation constant ; K[tex]_{a}[/tex] = [tex]\frac{(c\alpha)^{2} }{c-c\alpha }[/tex] ( HA ---> H⁺A⁻ )
∝ = [tex]\frac{2.1}{100}[/tex] = 0.021
[ H⁺ ] = c * ∝ = 0.115 * 0.021 = 0.0024
Given that
pH = -log [ H⁺ ]
pH = - log [ 0.0024 ]
∴ pH of the solution = 2.6
Hence we can conclude that the pH of the solution is 2.6
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