Hello!
datos: Molarity = [tex] 1.0*10^{-4}\:M\:(mol/L) [/tex]
ps: The ionization constant of the sulfuric acid is strong and completely dissociates in water, so the pH will be:
[tex] pH = - log\:[H_3O^+] [/tex]
[tex] pH = - log\:[1*10^{-4}] [/tex]
[tex] pH = 4 - log\:1 [/tex]
[tex] pH = 4 - 0 [/tex]
[tex] \boxed{\boxed{pH = 4}}\end{array}}\qquad\checkmark [/tex]
Note:. The pH <7, then we have an acidic solution.
I Hope this helps, greetings ... DexteR!
Answer:
The pH of the sulfuric acid solution is 3.70.
Explanation:
The pH is negative logarithm of hydrogen ion concentration.
[tex]pH=-\log[H^+][/tex]
Concentration of sulfuric acid = [tex]\times 10^{-4} M[/tex]
[tex]H_2SO_4(aq)\rightarrow 2H^+(aq)+SO_{4}^{2-}(aq)[/tex]
1 mole of sulfuric acid gives 2 moles of hydrogen ions.Then[tex]1.0\times 10^{-4} M[/tex] of sulfuric acid will give .
[tex][H^+]=2\times 1.0\times 10^{-4} M=2.0\times 10^{-4} M[/tex]
The pH of the solution :
[tex]pH=-\log[2.0\times 10^{-4} M]=3.70[/tex]
