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

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