# Calculate [H3O ], [ClO4], and [OH] in an aqueous solution that is 0.155 M in HClO4(aq) how [Solved] The pH of that is -log(0.155) = 0.81

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

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 = 110^-14
Due to this fact [HO-] = 110^-14 / 0.155 = 6.4510^-14

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 = 110^-14
Due to this fact [HO-] = 110^-14 / 0.155 = 6.4510^-14

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 = 110^-14
Due to this fact [HO-] = 110^-14 / 0.155 = 6.4510^-14

So you have got the focus of [H3O ] = [ClO4-] = 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 = 110^-14
Due to this fact [HO-] = 110^-14 / 0.155 = 6.4510^-14

Taking the adverse log of 110^-14 = [H3O ][OH-] yields 14 = pH pOH

HClO4 > H ClO4- (H = H3O ) Since HClO4 dissociates 100%, the focus of H and ClO- will each be 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 = 110^-14
Due to this fact [HO-] = 110^-14 / 0.155 = 6.4510^-14

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

This post is last updated on hrtanswers.com at Date : 1st of September – 2022