How do you know the ph level of 10^-8

pH=6.98

Explanation:

This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.

As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.

2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!

At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have

KW=[H3O+]⋅[OH−]=10−14

Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have

[H3O+]=√10−14=10−7M

The pH of pure water will thus be

pH=−log([H3O+])

pH=−log(10−7)=7

Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10