Question

If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4 is 2.40×10−5. express your answer to three significant figures and include the appropriate units.

Answer 1

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 $\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }$.

Question:

What mass of calcium sulfate will precipitate?

The Process:

Step-1

The balanced equilibrium reaction:

$\boxed{ \ CaSO_4_{(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)} \ }$ $\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }$

Let us prepare moles for both ions.

$\boxed{ \ M = \frac{n}{V} \ }$ $\rightarrow \boxed{ \ n = MV \ }$

$\boxed{ \ Moles \ of \ Ca^{2+} = 0.10 \ \frac{mole}{L} \times 0.5 \ L = 0.05 \ moles \ }$

$\boxed{ \ Moles \ of \ SO_4^{2+} = 0.10 \ \frac{mole}{L} \times 0.5 \ L = 0.05 \ moles \ }$

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.

$\boxed{ \ [Ca^{2+}] = \frac{0.05 \ moles}{1 \ L} = 0.05 \ M \ }$

$\boxed{ \ [SO_4^{2-}] = \frac{0.05 \ moles}{1 \ L} = 0.05 \ M \ }$

Step-2

Let us calculate the ion product (Q) and compare it to Ksp.

$\boxed{ \ Q = [Ca^{2+}][SO_{4}^{2-}] \ }$ $\rightarrow \boxed{ \ Q = [0.05][0.05] = 2.50 \times 10^{-3} \ M^2 \ }$

Compare with $\boxed{ \ K_{sp} = 2.40 \times 10^{-5} \ }$.

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

$\boxed{ \ K_{sp} = [Ca^{2+}][SO_{4}^{2-}] \ }$

$\boxed{ \ K_{sp} = s \times s \ }$

$\boxed{ \ K_{sp} = s^2 \ } \rightarrow \boxed{ \ s = \sqrt{K_{sp}} \ }$

$\boxed{ \ s = \sqrt{2.40 \times 10^{-5}} \ }$

Hence, the solubility of CaSO₄ is $\boxed{ \ 4.90 \times 10^{-3} \ M \ }$.

After that, we can find out the mole of CaSO₄ which is dissolved.

$\boxed{ \ Moles \ of \ CaSO_4 = 4.90 \times 10^{-3} \ \frac{mole}{L} \times (0.5 + 0.5) \ L = 4.90 \times 10^{-3} \ moles \ }$

Step-4

Thus we know that in this reaction:

$\boxed{ \ CaSO_4_{(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)} \ }$ 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 = $\boxed{ \ 0.05 \ moles - 4.90 \times 10^{-3} \ moles = 0.0451 \ moles \ }$

Final Step

Let us calculate the mass of CaSO₄ that will precipitate.

$\boxed{ \ n = \frac{mass}{Mr} \ }$ $\rightarrow \boxed{ \ mass = n \times Mr \ }$

The molar mass of CaSO₄ is 136 g/mole.

$\boxed{ \ mass = 0.0451 \ moles \times 136 \ \frac{g}{mole} \ }$

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.

Learn more

1. What is the Ksp of the salt at 22°C? brainly.com/question/8985555
2. How many grams of sodium hydroxide are needed to make 250 ml of a 7.80 M solution? brainly.com/question/12286318  
3. An example of the dilution of a solution brainly.com/question/4516437

Answer 2

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$.

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of  $SO_4^{2-}$, it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams