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