Lets write the dissociation equation of HClO
HClO -----> H+ + ClO-
0.41 0 0
0.41-x x x
Ka = [H+][ClO-]/[HClO]
Ka = x*x/(c-x)
Assuming small x approximation, that is lets assume that x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((3.5*10^-8)*0.41) = 1.198*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.198*10^-4 M
so.[H+] = x = 1.198*10^-4 M
we have below equation to be used:
pH = -log [H+]
= -log (1.198*10^-4)
= 3.92
Answer: 3.92
Feel free to comment below if you have any doubts or if this answer do not work
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