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3 years ago

Calculate the volume required to dissolve 11.2 g cuso4 to prepare a 0.140m solution

Chemistry

1 answer:

joja [24]3 years ago

4 0

Answer is: 0.5 dm³ to dissolve.

m(CuSO₄) = 11.2 g.

c(CuSO₄) = 0.140 mol/dm³,

V(CuSO₄) = ?

n(CuSO₄) = m(CuSO₄) ÷ M(CuSO₄).

n(CuSO₄) = 11.2 g ÷ 159.6 g/mol.

n(CuSO₄) = 0.07 mol.

V(CuSO₄) = n(CuSO₄) ÷ c(CuSO₄).

V(CuSO₄) = 0.07 mol ÷ 0.14 mol/dm³.

V(CuSO₄) = 0.5 dm³.

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Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

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At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

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