Answer:
[tex]pH=1.67[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the acid's pH, by firstly calculating the concentration of the hydrogen ions released by the acid in agreement to its dissociation in the 25% with a concentration of 0.085 N in both normality and molarity because it is monobasic:
[tex]0.25=\frac{[H^+]}{0.085M} \\\\[/tex]
[tex][H^+]=0.25*0.085M=0.02125M[/tex]
Next, we apply the logarithmic definition of pH to obtain:
[tex]pH=-log([H^+])\\\\pH=-log(0.02125M)\\\\pH=1.67[/tex]
Regards!
The pH of the 0.085 N monobasic acid which dissociates 25% has been 1.67.
The given acid is a monobasic acid. Thus, the normality and molarity are the same.
Molarity of the solution = 0.085 M
% Dissociation = [tex]\rm \dfrac{[H^+]}{molarity}[/tex]
0.25 = [tex]\rm \dfrac{[H^+]}{0.085}[/tex]
[tex]\rm [H^+][/tex] = 0.25 [tex]\times[/tex] 0.085
[tex]\rm [H^+][/tex] = 0.0212 M
The pH of the solution has been the logarithmic value of the hydrogen ion concentration.
pH = - log[tex]\rm [H^+][/tex]
pH = - log (0.0212)
pH = 1.67
The pH of the 0.085 N monobasic acid which dissociates 25% has been 1.67.
For more information about pH of the solution, refer to the link:
https://brainly.com/question/20437978
