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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.