Question

I have a few Chemistry questions, I have tried to go through them but keep getting stuck. Can anyone help me please?

Answer 1

The first two I have are precipitate questions. I am not sure whether the distilled water "topping" up the volume makes any difference to my calculations? for the first one I Keep getting 7.599g BaCrO4 which is the precipitate? But perhaps I am not factoring the amount that is used (if any in the reaction). So lost....

Answer 2

Where do you get the 0.06 from? Solved that from the equilibrium condition?

Answer 3

\[0.03 mol K2CrO4 (\frac{ 2 mol KCl }{ 1 mol K2CrO4 }) =0.06\]

Answer 4

He's referring to the concentration for KCl.

Answer 5

I used that 0.06 mol KCl and since the volume is 1L had concentration for KCl as 0.06M. Is this right?

Answer 6

if you had 100% conversion from potassium chromate, but you're not using the limiting reactant which is barium chloride.

Answer 7

So the most possible moles of barium chromate made was 0.02 moles.

Now how much remains dissolved and how much is solid?

The Ksp equation gives you the solubility.

So 1.45*10^(-5) M, the moles of barium chromate dissolved are

\(\sf M=\dfrac{moles}{L}\rightarrow moles=molarity*L=1.45*10^{-5} M*1L=1.45*10^{-5} moles\)

Now, we subtract whats dissolved from the total:

0.02-1.45*10^(-5)=0.0199855086232538 moles

Multiply by the molar mass of barium chromate, the mass is 5.0637 g

Answer 8

Thankyou. So just to clarify. As BaCl2 was the limiting that is why the BaCrO4 has to be 0.02 moles also? Then, the rest of the calculations I had done work. Thank you so much :)

Answer 9

I am stuck on another precipitation question but this time I need to calculate the concentration of the magnesium ions remaining in solution?

Answer 10

The 0.02 moles of BaCl2 and 0.02 moles of barium chromate would not exist simultaneously. BaCl2 breaks apart in solution and the barium ion finds chromate.

The most possible (theoretical) amount of barium chromate you could make is related to the moles of the limiting reactant, you have to factor out the coefficients. Have you ever done an I.C.E. table? Not that you need it, but it illustrates the concept.

Answer 11

I have done ICE tables but get mixed up with what is the change and what is initial etc

Answer 12

It's a similar concept. Write the Ksp and use the \(\sf [OH^-] \) obtained from the pH.

Answer 13

(that was for the second question)

Answer 14

Hm you should work with the ICE table concept, there are tons of examples on youtube, or we could do one here if you'd like - i think the visuals (youtube) are more effecitve.

Answer 15

cool, I will watch some and get my head around the concept :)

Answer 16

awesome! do ask again if you're still confused

Answer 17

Could you please help walk me through the PH Question?

Answer 18

So I put Ksp = [Na] [OH] and then use Na as X and OH as PH 12.0?

Answer 19

sure. now that i'm doing it, i dont think you need the pH at all - since it's only asking for the concentration remaining in solution - which is given by the Ksp.

The trick here is writing the Ksp equation, we need the balanced equation for the process, can you provide that?

Answer 20

ps. if you want to find \(\sf [OH^-]\) from pH, you need to find pOH first.

pH + pOH= 14 pOH=14-pH

\(\sf pOH=-log[OH^-]\)

Answer 21

I had NaFl + CaCl = CaFl +NaCl

Answer 22

Handy to remember :)

Answer 23

:)

Answer 24

um thats not the right equation :S

Answer 25

oh oops..

Answer 26

\(\sf NaOH + MgX_2 \rightleftharpoons Mg(OH)_2 +NaX\)

X is just a soluble counter-ion

Answer 27

can you balance it?

Answer 28

2Na + MgX2 = Mg(OH)2 +2NaX

Answer 29

2NaOH***

Answer 30

okay cool. Now we need the net ionic equation, can you do that?

Answer 31

Is it 2OH +Mg =Mg(OH)2 as the Na and X are the spectators?

Answer 32

yup, you need to show the charges of the ions.

the Ksp equation is: \(\sf K_{sp}=(2[OH^-])^2*[Mg^{2+}]\)

we let \(\sf [OH^-]=[Mg]^{2+}=x\)

then we have, \(\sf Ksp=(2x^2)*x=4x^3\)

Answer 33

I have to get going, you should repost this question if you still need help with it!

Answer 34

Thank you so much for your help :)

Answer 35

no problem :)