Calculate [h ], [clo4–], and [oh–] in an aqueous solution that is 0.175 m in hclo4(aq) at 25 °c.

HClO4(aq) ===> H+(aq) + ClO4-(aq)
If [HClO4] = 0.175 M, then [H+] = [ClO4-] = 0.175 M
[OH-] = 0.175 M = 1.75 x 10^-1 M
Kw = [H+][OH-} = 1.00 x 10^-14     [OH-] = 1.00x10^-4/1.75x10^-1 = 5.71 x 10^-14 M

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BarrettArcher BarrettArcher · Answer 2 · 2019-03-18T07:51:49+01:00

Answer : The concentration of [tex][H^+],[ClO_4^-]\text{ and }[OH^-][/tex] are, [tex]0.175M,0.175M\text{ and }5.70\times 10^{-14}M[/tex] respectively.

Solution : Given,

concentration of [tex]HClO_4[/tex] solution = 0.175 M

The balanced equilibrium reaction will be,

[tex]HClO_4\rightleftharpoons H^++ClO_4^-[/tex]

If the [tex]HClO_4[/tex] dissociates 100 percent then the concentration of [tex]H^+[/tex] and [tex]ClO_4^-[/tex] will be, 0.175 M

Thus,

[tex][H^+]=[ClO_4^-]=0.175M[/tex]

Now we have to calculate the pH.

[tex]pH=-\log [H^+]\\\\pH=-\log (0.175)=0.756[/tex]

Now we have to calculate the pOH.

[tex]pH+pOH=14\\\\pOH=14-pH=14-0.756=13.244[/tex]

Now we have to calculate the concentration of [tex]OH^-[/tex] ion.

[tex]pOH=-\log [OH^-]\\\\13.244=-\log [OH^-][/tex]

[tex][OH^-]=5.70\times 10^{-14}M[/tex]

Therefore, the concentration of [tex][H^+],[ClO_4^-]\text{ and }[OH^-][/tex] are, [tex]0.175M,0.175M\text{ and }5.70\times 10^{-14}M[/tex] respectively.