Respuesta :
Answer : The mass of calcium sulfate precipitate will be, 6.12 grams
Solution :
First we have to calculate the moles of [tex]Ca^{2+}[/tex] and [tex]SO_4^{2-}[/tex].
[tex]\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}[/tex]
[tex]\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}[/tex]
As, 0.05 moles of [tex]Ca^{2+}[/tex] is mixed with 0.05 moles of [tex]SO_4^{2-}[/tex], it gives 0.05 moles of calcium sulfate.
Now we have to calculate the solubility of calcium sulfate.
The balanced equilibrium reaction will be,
[tex]CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=(s)\times (s)[/tex]
[tex]K_{sp}=(s)^2[/tex]
Now put the value of [tex]K_{sp}[/tex] in this expression, we get the solubility of calcium sulfate.
[tex]2.40\times 10^{-5}=(s)^2[/tex]
[tex]s=4.89\times 10^{-3}M[/tex]
Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.
[tex]\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}[/tex]
Now we have to calculate the moles of calcium sulfate that precipitated.
[tex]\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}[/tex]
[tex]\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}[/tex]
Now we have to calculate the mass of calcium sulfate that precipitated.
[tex]\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4[/tex]
[tex]\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g[/tex]
Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams
6.13 grams
Further explanation
Given:
- 500.0 ml of 0.10 M Ca²⁺ is mixed with 500.0 ml of 0.10 M SO₄²⁻.
- Ksp for CaSO₄ is [tex]\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }[/tex].
Question:
What mass of calcium sulfate will precipitate?
The Process:
Step-1
The balanced equilibrium reaction:
[tex]\boxed{ \ CaSO_4_{(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)} \ }[/tex] [tex]\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }[/tex]
Let us prepare moles for both ions.
[tex]\boxed{ \ M = \frac{n}{V} \ }[/tex] [tex]\rightarrow \boxed{ \ n = MV \ }[/tex]
[tex]\boxed{ \ Moles \ of \ Ca^{2+} = 0.10 \ \frac{mole}{L} \times 0.5 \ L = 0.05 \ moles \ }[/tex]
[tex]\boxed{ \ Moles \ of \ SO_4^{2+} = 0.10 \ \frac{mole}{L} \times 0.5 \ L = 0.05 \ moles \ }[/tex]
Then prepare the concentration of each ion after mixing. Remember, after mixing we get a total volume of 500 mL + 500 mL = 1,000 mL or 1 L.
[tex]\boxed{ \ [Ca^{2+}] = \frac{0.05 \ moles}{1 \ L} = 0.05 \ M \ }[/tex]
[tex]\boxed{ \ [SO_4^{2-}] = \frac{0.05 \ moles}{1 \ L} = 0.05 \ M \ }[/tex]
Step-2
Let us calculate the ion product (Q) and compare it to Ksp.
[tex]\boxed{ \ Q = [Ca^{2+}][SO_{4}^{2-}] \ }[/tex] [tex]\rightarrow \boxed{ \ Q = [0.05][0.05] = 2.50 \times 10^{-3} \ M^2 \ }[/tex]
Compare with [tex]\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }[/tex].
Because Q > Ksp, the solution is supersaturated and CaSO₄ will precipitate from solution.
Step-3
Let us calculate the solubility of CaSO₄ (s).
CaSO₄ ⇄ Ca²⁺ + SO₄²⁻
s s s
[tex]\boxed{ \ K_{sp} = [Ca^{2+}][SO_{4}^{2-}] \ }[/tex]
[tex]\boxed{ \ K_{sp} = s \times s \ }[/tex]
[tex]\boxed{ \ K_{sp} = s^2 \ } \rightarrow \boxed{ \ s = \sqrt{K_{sp}} \ }[/tex]
[tex]\boxed{ \ s = \sqrt{2.40 \times 10^{-5}} \ }[/tex]
Hence, the solubility of CaSO₄ is [tex]\boxed{ \ 4.90 \times 10^{-3} \ M \ }[/tex].
After that, we can find out the mole of CaSO₄ which is dissolved.
[tex]\boxed{ \ Moles \ of \ CaSO_4 = 4.90 \times 10^{-3} \ \frac{mole}{L} \times (0.5 + 0.5) \ L = 4.90 \times 10^{-3} \ moles \ }[/tex]
Step-4
Thus we know that in this reaction:
[tex]\boxed{ \ CaSO_4_{(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)} \ }[/tex] 0.05 moles of Ca²⁺ mixed with 0.05 moles of SO₄²⁻, will produce 0.05 moles of CaSO₄.
To calculate the precipitated mole of CaSO₄, we must subtract the resulting CaSO₄ mole with the dissolved CaSO₄ mole.
Moles of CaSO₄ precipitated = [tex]\boxed{ \ 0.05 \ moles - 4.90 \times 10^{-3} \ moles = 0.0451 \ moles \ }[/tex]
Final Step
Let us calculate the mass of CaSO₄ that will precipitate.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ }[/tex] [tex]\rightarrow \boxed{ \ mass = n \times Mr \ }[/tex]
The molar mass of CaSO₄ is 136 g/mole.
[tex]\boxed{ \ mass = 0.0451 \ moles \times 136 \ \frac{g}{mole} \ }[/tex]
Thus, the mass of calcium sulfate which will precipitate by 6.13 grams.
_ _ _ _ _ _ _ _ _ _
Notes
- If Q > Ksp, the solution is supersaturated. Ion concentrations > equilibrium concentrations, the reaction will proceed in reverse to reach equilibrium, precipitation will occur.
- If Q < Ksp, the solution is unsaturated. Ion concentrations < equilibrium concentrations, the reaction will proceed forward to reach equilibrium, more solid will dissolve.
- If Q = Ksp, the solution is saturated. The reaction rate goes both ways with the same value. Ion concentrations = equilibrium concentrations, no more solid will dissolve or precipitate.
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