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

Answer:

The pH of the sulfuric acid solution is 3.70.

Explanation:

The pH is negative logarithm of hydrogen ion concentration.

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

Concentration of sulfuric acid = $\times 10^{-4} M$

$H_2SO_4(aq)\rightarrow 2H^+(aq)+SO_{4}^{2-}(aq)$

1 mole of sulfuric acid gives 2 moles of hydrogen ions.Then$1.0\times 10^{-4} M$ of sulfuric acid will give .

$[H^+]=2\times 1.0\times 10^{-4} M=2.0\times 10^{-4} M$

The pH of the solution :

$pH=-\log[2.0\times 10^{-4} M]=3.70$