Question

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

Answer 1

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

Answer 2

Answer : The concentration of
`[H^+],[ClO_4^-]\text{ and }[OH^-]` are,
`0.175M,0.175M\text{ and }5.70* 10^(-14)M` respectively.

Solution : Given,

concentration of
`HClO_4` solution = 0.175 M

The balanced equilibrium reaction will be,

`HClO_4\rightleftharpoons H^++ClO_4^-`

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

Thus,

`[H^+]=[ClO_4^-]=0.175M`

Now we have to calculate the pH.

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

Now we have to calculate the pOH.

`pH+pOH=14\\\\pOH=14-pH=14-0.756=13.244`

Now we have to calculate the concentration of
`OH^-` ion.

`pOH=-\log [OH^-]\\\\13.244=-\log [OH^-]`

`[OH^-]=5.70* 10^(-14)M`

Therefore, the concentration of
`[H^+],[ClO_4^-]\text{ and }[OH^-]` are,
`0.175M,0.175M\text{ and }5.70* 10^(-14)M` respectively.