Question

Question 8 (5 points)

Answer 1

[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