Respuesta :

First calculate the Molarity of the final solution :

C
₁ = 6.0 M

V₁ = 25 mL

C₂ = ?

V₂ = 1.75 * 1000 = 1750 mL

C₁ * V₁ = C₂ * V₂

6.0 * 25 = C₂ * 1750

150 = C₂ * 1750

C₂ = 150 / 1750

C₂ = 0.08571 M

Therefore:

pH = - log [C₂]

pH = - log [ 0.08571 ]

pH = 1.0669

hope this helps!


Answer:

could someone comment the pOH of this solution?

Explanation:

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