Respuesta :
Explanation:
Ionization equation for [tex]CaSO_{4}[/tex] is as follows.
[tex]CaSO_{4} \rightarrow Ca^{2+} + SO^{2-}_{4}[/tex]
s s s
Now, the expression for the solubility product is as follows.
[tex]K_{sp} = [Ca^{2+}][SO^{2-}_{4}][/tex]
= [tex]s \times s[/tex]
= [tex]s^{2}[/tex]
As the concentration of [tex]Na_{2}SO_{4}[/tex] is given as 0.4 M.
So, [tex][Na_{2}SO_{4}] = [SO^{2-}_{4}][/tex] = 0.4 M
Putting the given values as follows.
[tex]K_{sp} = [Ca^{2+}][SO^{2-}_{4}][/tex]
[tex]4.93 \times 10^{-5} = [Ca^{2+}] \times 0.4[/tex]
[tex][Ca^{2+}][/tex] = 12.325 \times 10^{-5}[/tex]
Hence, the solubility of [tex]CaSO_{4}[/tex] in [tex]Na_{2}SO_{4}[/tex] is [tex]12.325 \times 10^{-5}[/tex].
Therefore, solubility of [tex]CaSO_{4}[/tex] in g/ml as follows.
[tex]12.325 \times 10^{-5} \times 136 g/mol[/tex]
= 0.0167 g/L
Thus, we can conclude that solubility of [tex]CaSO_{4}[/tex] is 0.0167 g/L.
The solubility of CaSO₄ in 0.4M of Na₂SO₄ is given as;
0.016762 g/mol
We are given;
Solubility of CaSO₄; K_sp = 4.93 × 10⁻⁵
Concentration of Na₂SO₄ = 0.4 M
The ionic equation of CaSO₄ is given as;
CaSO₄ ⇒ Ca²⁺ + SO₄²⁻
Now, the formula for solubility is;
K_sp = [A⁺]ᵃ × [B⁻]ᵇ
where;
A⁺ = Cations present in the aqueous solution
B⁻ = Anions present in the aqueous solution
a and b are relative concentrations
Thus, plugging in relevant values into the solubility equation, we have;
4.93 × 10⁻⁵ = Ca²⁺ × 0.4
Ca²⁺ = ( 4.93 × 10⁻⁵)/0.4
Concentration of Ca²⁺ = 12.325 × 10⁻⁵ M
Now, the molar mass of CaSO₄ is 136 g/mol
Thus, solubility of CaSO₄ in 0.4M of Na₂SO₄ is;
12.325 × 10⁻⁵ M × 136 g/mol = 0.016762 g/mol
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