[H+][OH-]=10-¹⁴
We are using this formula because we need to find the H+
substitute the value given for hydronium ion for OH-
[H+][4.19×10⁵]=10-¹⁴
[H+]=10-¹⁴÷4.19×10⁵
[H+]=2.387×10-¹⁹
Then the pH of the solution will be
pH= –log¹⁰ [H+]
pH = –log¹⁰ [2.387×10-¹⁹]
pH= –log¹⁰2.387+19log¹⁰
= –0.378+19
pH =18.622
