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kompoz [17]
2 years ago
13

Ion-dipole interactions can occur between any ion and any molecule with a dipole. Identify all of the following pairs of species

that can interact via ion-dipole forces. chegg
Chemistry
1 answer:
Zigmanuir [339]2 years ago
4 0

An ion-dipole interaction is the result of an electrostatic interaction between a charged ion and a molecule that has a dipole. It is an attractive force that is commonly found in solutions, especially ionic compounds dissolved in polar liquids. A cation can attract the partially negative end of a neutral polar molecule, while an anion attracts the positive end of a polar molecule. Ion-dipole attractions become stronger as the charge on the ion increases or as the magnitude of the dipole of the polar molecule increases.

This force of attraction is between an ion and a charge , it is weaker force than covalent bond and ionic bond . EX - The ion dipole interaction takes place between water and sodium ion , in it there is a small charge on oxygen molecule in water which is attracted by sodium charge .

Most commonly found in solutions. Especially important for solutions of ionic compounds in polar liquids.

A positive ion (cation) attracts the partially negative end of a neutral polar molecule.

to learn more about dipole interactions:-

https://brainly.in/question/1157107

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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
attashe74 [19]

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

5 0
3 years ago
Read 2 more answers
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