Question

Calculate the [OH-] and the pH of 0.035 M KOH.

Answer 1

pH

`= 12.54`

OH- concentration 28.84

Step-by-step explanation:

KOH dissociates into K+ and OH-. The ratio of K+ and OH- ion is 1:1

In any aqueous solution, the H3O+ and OH - must satisfy the following condition -

`[ H_3O^+] [OH^-] = k_w`

`[ H_3O^+] = (k_w)/( [OH^-])`

M

pH =

`- log [ H_3O^+]\\- log [2.857 * 10^(-13)]`

pH

`= - [-12.54]\\= 12.54`

pOH
`= 14 -` pH

pOH

`= 14 - 12.54\\= 1.46`

OH concentration
`= antilog 1.46`

OH- concentration 28.84