How do we do this? (1 Viewer)

[Cherrybomb56]

Can someone please explain both these questions to me?

 

[Qeru]

Can someone please explain both these questions to me?

9. Draw up an ICE table.

	NOCl	NO	Cl
I	1,6	0	0
C	-0.6	0.6	0.3
E	1	0.6	0.3

The change for NOCl and NO are the same whereas the same for Cl is half that of NOCl since the molar ratio is 2:2:1. So in total 1.9 moles.

 

[black.mamba]

Q18)

Bonds Broken:

• Ionic bonds (between sodium cations and chloride anions) are broken
• Hydrogen bonding between water molecules is disrupted (minor factor, you probably don't need to mention this)

Bonds Formed:

• Ion-dipole bonds form between the dipoles of water molecules and the ions in solution

Entropy:

• Entropy usually increases during the dissolution of an ionic solid due to the increased disorder of the system once the ions dissociate
• As NaCl is highly soluble, the Gibb's energy of dissolution is negative, which is due to the increase in entropy, so in this case, entropy does increase significantly

Diagrams:

• Draw a diagram before dissolution (ionic lattice and water molecules surrounding it)
• Draw a diagram during dissolution (ion is pulled away from lattice by water molecules)
• Draw a diagram after dissolution (lone ion surrounded by water molecules)

For the diagrams, make sure that the correct dipole of water is facing the ion (e.g. the negative dipole of water, or the oxygen side, faces the cation). Also, label the ions and draw in charges (for the ions) and partial charges (for the dipoles), and draw in the ion-dipole bonds.

 

[Qeru]

Can someone please explain both these questions to me?

18 is jsut copied from the nesa sample questions:

Here is a sample response (but for KCl)

 

[CM_Tutor]

Q18)

Bonds Broken:

• Ionic bonds (between sodium cations and chloride anions) are broken
• Hydrogen bonding between water molecules is disrupted (minor factor, you probably don't need to mention this)

Bonds Formed:

• Ion-dipole bonds form between the dipoles of water molecules and the ions in solution

Entropy:

• Entropy usually increases during the dissolution of an ionic solid due to the increased disorder of the system once the ions dissociate
• As NaCl is highly soluble, the Gibb's energy of dissolution is negative, which is due to the increase in entropy, so in this case, entropy does increase significantly

Diagrams:

• Draw a diagram before dissolution (ionic lattice and water molecules surrounding it)
• Draw a diagram during dissolution (ion is pulled away from lattice by water molecules)
• Draw a diagram after dissolution (lone ion surrounded by water molecules)

For the diagrams, make sure that the correct dipole of water is facing the ion (e.g. the negative dipole of water, or the oxygen side, faces the cation). Also, label the ions and draw in charges (for the ions) and partial charges (for the dipoles), and draw in the ion-dipole bonds.

There's only one point that I would add to this...

The enthalpy of dissolution is made up of two competing parts:

• lattice energy (endothermic) involved in breaking apart the ionic lattice and separating ions from each other
• hydration enthalpy (exothermic) involved in forming the ion-dipole interactions

The process of dissolution begins at corners where there are fewer ions / ionic bonds holding a particular ion in place because there is then good access for the formation of the ion-dipole interactions to pull the ion into solution. This is an entropically-favourable process.

The process can be overall exothermic (as in something like NaOH) or endothermic (as in something like KNO₃) but in either case can be spontaneous () on entropic grounds.

 

[Cherrybomb56]

Thanks for the help @Qeru, @black.mamba , and @CM_Tutor. Could anyone also please explain this to me?

 

[black.mamba]

Pb(NO3)2(aq) + 2KCl(aq) = PbCl2(s) + 2KNO3(aq)

(using equals as the double-headed arrows)

In the above equation, nitrate and potassium are spectators ions (i.e. they are aqueous throughout). We can remove them to get the net ionic equation:

Pb2+(aq) + 2Cl-(aq) = PbCl2(s)

(sorry for the shitty formatting)

From the question, we know:
n(Pb2+) = n(Pb(NO3)2) = 0.001 mol
n(PbCl2) = n (Pb2+) = 0.001 mol

Accounting for the 25mL volume,
[PbCl2] = n(PbCl2)/v = 0.001/0.025 = 1/25 mol/L = 0.04 mol/L

The solubility of lead (II) chloride is given as 1.0g/100g of water (which is equal to 10g/L)
For m(PbCl2) = 10, n = m/MM = 10/278.1 = 0.036 mol (approx)
So the solubility can be expressed as 0.036 mol/L

Because the concentration is greater than 0.036 mol/L, therefore a precipitate will form

 

[CM_Tutor]

Pb(NO3)2(aq) + 2KCl(aq) = PbCl2(s) + 2KNO3(aq)

(using equals as the double-headed arrows)

In the above equation, nitrate and potassium are spectators ions (i.e. they are aqueous throughout). We can remove them to get the net ionic equation:

Pb2+(aq) + 2Cl-(aq) = PbCl2(s)

(sorry for the shitty formatting)

From the question, we know:
n(Pb2+) = n(Pb(NO3)2) = 0.001 mol
n(PbCl2) = n (Pb2+) = 0.001 mol

Accounting for the 25mL volume,
[PbCl2] = n(PbCl2)/v = 0.001/0.025 = 1/25 mol/L = 0.04 mol/L

The solubility of lead (II) chloride is given as 1.0g/100g of water (which is equal to 10g/L)
For m(PbCl2) = 10, n = m/MM = 10/278.1 = 0.036 mol (approx)
So the solubility can be expressed as 0.036 mol/L

Because the concentration is greater than 0.036 mol/L, therefore a precipitate will form

But there are two different solutions of 25 mL, so the total volume in the ultimate solution is 50 mL...

 

[black.mamba]

But there are two different solutions of 25 mL, so the total volume in the ultimate solution is 50 mL...

Oh yikes I’m rusty already my bad

 

[black.mamba]

Pb(NO3)2(aq) + 2KCl(aq) = PbCl2(s) + 2KNO3(aq)

(using equals as the double-headed arrows)

In the above equation, nitrate and potassium are spectators ions (i.e. they are aqueous throughout). We can remove them to get the net ionic equation:

Pb2+(aq) + 2Cl-(aq) = PbCl2(s)

(sorry for the shitty formatting)

From the question, we know:
n(Pb2+) = n(Pb(NO3)2) = 0.001 mol
n(PbCl2) = n (Pb2+) = 0.001 mol

Accounting for the 25mL volume,
[PbCl2] = n(PbCl2)/v = 0.001/0.025 = 1/25 mol/L = 0.04 mol/L

The solubility of lead (II) chloride is given as 1.0g/100g of water (which is equal to 10g/L)
For m(PbCl2) = 10, n = m/MM = 10/278.1 = 0.036 mol (approx)
So the solubility can be expressed as 0.036 mol/L

Because the concentration is greater than 0.036 mol/L, therefore a precipitate will form

Nvm halve the concentration above so it’s actually 0.020 mol/L and there won’t be a precipitate

 

[CM_Tutor]

Oh yikes I’m rusty already my bad

We all make mistakes. To be clear, I pointed out the error so that the OP would be aware of it and correct it (or look at your subsequent correction).

In my own teaching, I encourage students to ask questions / point out problems if they see an error (or possible error) that I have made. I view those moments as opportunities to model reflection and valuable metacognitive skills, as moments to reinforce that everyone makes mistakes, and to encourage questioning when we don't understand. Sometimes the error is not a mistake but reveals a way in which an explanation has been deficient. Sometimes (though not here at BoS), I make a deliberate error both to see if it is caught and to create an opportunity for reinforcing a teaching point, or to model the skill of identifying an error when it is known to exist (such as when a Maths proof gets the wrong answer, or a Chemistry calculation leads to an impossible conclusion).

Short version: Black.mamba, I hope you didn't feel that I was trying to embarrass you or anything like that.

 

[black.mamba]

We all make mistakes. To be clear, I pointed out the error so that the OP would be aware of it and correct it (or look at your subsequent correction).

In my own teaching, I encourage students to ask questions / point out problems if they see an error (or possible error) that I have made. I view those moments as opportunities to model reflection and valuable metacognitive skills, as moments to reinforce that everyone makes mistakes, and to encourage questioning when we don't understand. Sometimes the error is not a mistake but reveals a way in which an explanation has been deficient. Sometimes (though not here at BoS), I make a deliberate error both to see if it is caught and to create an opportunity for reinforcing a teaching point, or to model the skill of identifying an error when it is known to exist (such as when a Maths proof gets the wrong answer, or a Chemistry calculation leads to an impossible conclusion).

Short version: Black.mamba, I hope you didn't feel that I was trying to embarrass you or anything like that.

ty for pointing out the mistake, it's helpful for the OP!