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d1i1m1o1n [39]
3 years ago
13

Someone help me choose three for extra points!

Chemistry
1 answer:
Lynna [10]3 years ago
3 0

question 2

a

b

d

question 3

1. b

2. c

3.a

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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
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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

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