Question

If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?

Answer 1

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

    Mn(OH)₂   →    Mn⁺² + 2OH⁻

    [Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

    Ksp = [Mn⁺²][OH-]²

    [Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

    [OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

    [OH⁻]² = 10⁻³

    [OH⁻] = (10⁻³)¹⁽²

    [OH⁻] = 0.0316 M

Calculate the pOH

    pOH = -log [OH⁻]

    pOH =  -log [0.0316]

    pOH = 1.5

Now calculate pH

   pH = 14 - pOH

   pH = 14 - 1.5

   pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4