Calculate and compare the [h+] and ph for a 0.100 m solution of hclo4 and a 0.100 m solution of hclo (ka = 2.9 × 10–8).

[H+] for HClO(this is a weak acid so it requires an I.C.E. table to solve):

(I=initial amount. C=change in amount. E= amount at equilibrium.)
I.C.E. table
HClO. H2O. >>> H3O+ ClO4-
I. 0.100M. N/A. 0. 0
C. -x. N/A. +x. +x
E. 0.1-x. N/A. x. x.
(we don't consider water because it is not an aqueous solution)

ka=(x^2)/(0.1-x)
(2.9×10^-8)= (x^2)/(0.1-x)
(0.1-x)(2.9×10^-8) = x^2
(3.0×10^-9) - ((2.9×10^-8)x) = x^2

x^2+(2.9×10^-8)x-(3×10^-9)

a=1
b=(2.9×10^-8)
c= -(3×10^-9)

plug those values into the quadratic formula:

x= (-b +(√((b^2)-4ac/2a

I got x= 0.000053837

x= [H+] =0.000053837

[H+] for 0.100 M HClO4:
because HClO4 is a strong acid, it dissociates completely. Meaning that it's [H+] =0.1

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

2.00 is the !