Question

What is the pH of a 1.0 x 10^-4 M solution of sulfuric acid (H2SO4)?

Answer 1

Hello!

datos: Molarity =
`1.0*10^(-4)\:M\:(mol/L)`

ps: The ionization constant of the sulfuric acid is strong and completely dissociates in water, so the pH will be:

`pH = - log\:[H_3O^+]`

`pH = - log\:[1*10^(-4)]`

`pH = 4 - log\:1`

`pH = 4 - 0`

`\boxed{\boxed{pH = 4}}\end{array}}\qquad\checkmark`

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR!

Answer 2

The pH of the sulfuric acid solution is 3.70.

Step-by-step explanation:

The pH is negative logarithm of hydrogen ion concentration.

`pH=-\log[H^+]`

Concentration of sulfuric acid =
`* 10^(-4) M`

`H_2SO_4(aq)\rightarrow 2H^+(aq)+SO_(4)^(2-)(aq)`

1 mole of sulfuric acid gives 2 moles of hydrogen ions.Then
`1.0* 10^(-4) M` of sulfuric acid will give .

`[H^+]=2* 1.0* 10^(-4) M=2.0* 10^(-4) M`

The pH of the solution :

`pH=-\log[2.0* 10^(-4) M]=3.70`