Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.155 M in HClO4(aq) – how?

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous resolution that’s 0.155 M in HClO4(aq) at 25 °C.

HClO4 —————–> H+ + ClO4- (H+ = H3O+) Since HClO4 dissociates 100%, the focus of H+ and ClO- will each be 0.155 mol/L

So you have got the focus of [H3O+] = [ClO4-] = 0.155 mol/L

The pH of that is -log(0.155) = 0.81

pH + pOH = 14

So the pOH = 14 – 0.81 = 13.19

[OH-] = 10^-13.19 = 6.45 x 10^-14

Al is appropriate:

“HClO4 —————–> H+ + ClO4- (H+ = H3O+) Since HClO4 dissociates 100%, the focus of H+ and ClO- will each be 0.155 mol/L

So you have got the focus of [H3O+] = [ClO4-] = 0.155 mol/L”

It is best to understand that the focus of OH^-1 might be calculated from Kw.
Kw = [H3O+][OH-]
At 25 deg C the worth of Kw = 1×10^-14
Due to this fact [HO-] = 1×10^-14 / 0.155 = 6.45×10^-14

Taking the adverse log of 1×10^-14 = [H3O+][OH-] yields 14 = pH + pOH

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