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[jackerino]

Question 25 (2 marks) Marks
Equal volumes of lead nitrate solution and sodium iodide solution, each with a concentration
of 0.10mol L-1, are mixed. A yellow precipitate of lead iodide results.

Calculate the concentration of each ion remaining in solution. 2
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[jackerino]

Question 22 (7 marks)
Marks
A chemistry student is required to produce 25.0 mL of 0.01M sodium carbonate
solution.
She was provided with the following forms of sodium carbonate:
• A bottle containing 200.0 mL of 0.05M sodium carbonate solution.
• A container of solid sodium carbonate.
Describe TWO methods the student could use to produce the required solution,
each using a different starting form of sodium carbonate provided.
Include calculations, the necessary equipment and the procedure to be followed in
your answer.
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[jackerino]

Omg help meee pleeeasse~!!!!!

 

[theind1996]

Dude, where'd you find this question? Q22.

This was in one of my school's past papers...

 

[jackerino]

Dude, where'd you find this question? Q22.

This was in one of my school's past papers...

I have friends in high places... lol

 

[Rawf]

I'm just guessing.. as I haven't done any other question like q25 but... (someone please correct me if I'm wrong)
Pb(NO3)4 + 4NaI --> PbI4 + 4NaNO3

Mol ratio of Pb(NO3)4:NaI is 1:4
As both solutions are 0.1mol/L each, you add the same amount of moles (equal volumes are added)

Therefore, assume we're adding 1L of both
0.1mol of Pb(NO3)4 is added and 0.1mol of NaI is added
Therefore... (from mol ratio) Pb(NO3)4 will be in excess (as you need x4 this amount for NaI to have a complete reaction)
0.1/4 mols = 0.025mols of Pb(NO3)4 is used, therefore 0.1-0.025 = 0.075mols is in excess --> Therefore conc of Pb(NO3)4 is 0.075mol/L

As 0.1mol of NaI are going to be used up, the final conc of NaI = 0mol/L

Mol ratio of Pb(NO3)4bI4 is 1:1
Therefore, as 0.025mols of Pb(NO3)4 is used --> The conc of PbI4 is 0.025mol/L (im not sure if we're supposed to add up all the volumes or what...)

Mol ratio of NaI:NaNO3 is 1:1
As 0.1mols of NaI is used --> The conc of NaNO3 is 0.1mol/L

So finally, conc of Pb(NO3)4 = 0.075mol/L
NaI = 0mol/L
PbI4 = 0.025mol/L
NaNO3 = 0.1mol/L
These concentrations might be wrong.. I'm not sure if you're supposed to divide them by 2 because the volume is 2L when you add the 2 solutions together.

For Q22, I'm not really gonna go through it in detail (not sure about the 2nd method though).
Method 1: (Using A container of solid sodium carbonate) - List the equipment and everything needed blablabla
1. Weigh 0.0265g of Na2CO3 solid on an electric balance
n= c x v = 2.5x10^-4
mass = n(molar mass) = 0.0265g
2. Transfer the solid to a 50mL or 100mL beaker
3. Using a 100mL measuring cylinder, add 25mL of distilled water to the beaker
4. Using a stirring rod, stir the solution until the solid is dissolved
(I'm not sure if this is the exact method lolol, but you can figure out how much moles of Na2CO3 you need in 250mL to make a solution of 0.1mol/L & find the mass, and then add it to a 250mL volumetric flask, fill it up with distilled water until the bottom of the water touches the meniscus, shake it and then use a 25mL pipette to take 25mL of this solution and place it in a separate beaker)

Method 2: (A bottle containing 200.0 mL of 0.05M sodium carbonate solution) - sorry idk if this method is right (calculation wise)
Using the calculation C1xV1 = C2xV2 we can figure out how much distilled water to add to the 0.05mol/L of Na2CO3 solution
0.05 x 0.2 = 0.01 x V2
Thus, V2 = 1L... so we add in 0.8L of distilled water
1. Pour the entire bottle of (200mL) of 0.05M Na2CO3 solution into a large volumetric flask (2L volumetric flask - lol idk if they have them?)
2. Add 0.8L of distilled water to the flask until the bottom of the distilled water touches the meniscus
3. Put the lid on the solution and shake it
4. Use a 25mL pipette and place 25mL of this solution into a separate beaker

Hope this helped