Respuesta :

wuzzleluver
[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
dsdrajlin

HClO₄ has a higher [H⁺] than HClO (0.100 M vs. 5.4 × 10⁻⁵ M).

HClO₄ has a lower pH than HClO (1.00 vs. 4.3).

Classification of acids according to their strength

  • Weak acids: dissociate partially in water.
  • Strong acids: dissociate completely in water.

HClO₄ is a strong acid. Thus, of HClO₄ is 0.100 M, H⁺ will be 0.100 M as well. We can use this value to calculate the pH for this acid.

pH = -log [H⁺] = -log 0.100 = 1.00

HClO is a weak acid. Thus, [H⁺] ≠ [HClO]. Given the acid dissociation constant (Ka) and the concentration of the acid (Ca), we can calculate [H⁺]  and pH using the following expressions.

[tex][H^{+} ] = \sqrt{Ca \times Ka } = \sqrt{0.100 \times (2.9 \times 10^{-8} ) } = 5.4 \times 10^{-5} \\\\pH = -log 5.4 \times 10^{-5} = 4.3[/tex]

HClO₄ has a higher [H⁺] than HClO (0.100 M vs. 5.4 × 10⁻⁵ M).

HClO₄ has a lower pH than HClO (1.00 vs. 4.3).

Learn more about acid strength here: https://brainly.com/question/14115968

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