NEED HELP WITH : SOLUBILITY

[boxcon]

ok i need some help i been doing one of my last cores for my grade 12 online and i been running in to some problem most of the time i can think my way out of it but some time i would like a little help. now you would say well go ask the teacher and i have the problem is he never gets back to me he allways just marks my stuff and i get full marks idk if im doing stuff right or if the way i figured it out works all the time because he never msg is me back i even emailed him and nothing soo i turn to you guys the science forums for help. soo i was wondering if you could tell me if i was doing a few of theys questions right and if i need to be doing something different

first up: 1. A saturated solution of BaSO4 is given to patients needing digestive tract x-rays.
a) Write an equation that represents the solubility equilibrium. (1 mark)
b) Calculate the [ba 2+] present in the saturated solution. (2 marks)

and this is my assure

1) A. BaSO4 <- -> Ba + SO4

B. ksp(BaSO4) = 1.1*10^-10= Ba^2 – SO4= S^2 S=/ 1.1*10^-10 =1.05*10^-8

S=1.05*10^-8

Ba = 1.05*10^-8 (M/L)

i had some problems with this and i dont feel that i have done it right but im not sure could you give me some pointers???

second:A saturated solution of AgCH3 COO was evaporated to dryness. The 250.0 mL sample was found to
contain 1.84 g AgCH3 COO. Calculate the solubility product constant for AgCH3 COO.

now this one i didnt use the 250.0 mL sample in any of the calculations soo i dont think i got it right but here what i did

4. AgCH3COO <- -> (AgCH3) (CO2)
MM= 166.9 = MM= 122.9 + MM=44

1.84g/166.9g * 1 mole=0.0110245656=0.0110 (conversion factor)

AgCH3=0.0110 CO2=0.0110 (using mole ratio)

Ksp(AgCH3COO)=0.0220

third:Will a precipitate form when 90.0 mL of 1. 00 × 10 −2 M Cu(NO3)2 and 10.0 mL
of 1. 00 × 10 −2 M NaIO3 are mixed? Explain using appropriate calculations. (3 marks)

i think i did this one right but want to make sure.

A. Ksp(Cu(IO3)2)= 6.9*10^-8 Cu(IO3)2 <-  Cu2 + IO3 90.0 mL of 1. 00 10 ^2 M CuNO32.

10.0 mL of 1.00*10^2 M NaIO3.

90.0ml+10.0ml= 100(FV) Cu2(FC)=1.00*10^-2 * 90/100 = 0.9*10^-3. IO3(FC) =1.00*10^2 *
10/100 =10 tiralKsp= (0.9*10^-3) *(10)= 0.9*10^-3

trialKsp(0.9*10^-3) < Ksp(6.9*10^-8 )

a precipitate will not form

think i have to change thins to will form thats about it.

any help would be greatly appreciated.

[boxcon]

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Edited May 22, 2014 by boxcon

[hypervalent_iodine]

first up: 1. A saturated solution of BaSO4 is given to patients needing digestive tract x-rays.

a) Write an equation that represents the solubility equilibrium. (1 mark)

b) Calculate the [ba 2+] present in the saturated solution. (2 marks)

 

and this is my assure

 

1) A. BaSO4 <- -> Ba + SO4

 

B. ksp(BaSO4) = 1.1*10^-10= Ba^2 – SO4= S^2 S=/ 1.1*10^-10 =1.05*10^-8

 

S=1.05*10^-8

 

Ba = 1.05*10^-8 (M/L)

 

i had some problems with this and i dont feel that i have done it right but im not sure could you give me some pointers???

 

 

The general form for Ksp is the same for any equilibria expression:

 

A_aB_b(s) <-->aA⁺ + bB^-

 

Ksp = [A+]^a[b-]^b

 

In your expression for Ksp for BaSO₄, you appear to have squared the concentration for Ba²⁺ ions when you shouldn't have. It also looks like you are subtracting [sO₄^2-] from [ba²⁺]? That is definitely incorrect. The concentrations should be multiplied. The second part looks correct.

 

 

second:A saturated solution of AgCH3 COO was evaporated to dryness. The 250.0 mL sample was found to

contain 1.84 g AgCH3 COO. Calculate the solubility product constant for AgCH3 COO.

 

now this one i didnt use the 250.0 mL sample in any of the calculations soo i dont think i got it right but here what i did

 

4. AgCH3COO <- -> (AgCH3) (CO2)

MM= 166.9 = MM= 122.9 + MM=44

 

1.84g/166.9g * 1 mole=0.0110245656=0.0110 (conversion factor)

 

AgCH3=0.0110 CO2=0.0110 (using mole ratio)

 

Ksp(AgCH3COO)=0.0220

Nope. I have a feeling you got the equation wrong because of the way that you wrote the question, but CH₃COO^- is the anionic component, not COO^-. When separating the ions, usually you would separate them into metal + nonmetal. You've also missed the fact that all equilibrium expressions work in terms of concentration, not number of moles. In other words, the number of moles in the 250 mL is not the same as the concentration of the solution.

 

 

third:Will a precipitate form when 90.0 mL of 1. 00 × 10 −2 M Cu(NO3)2 and 10.0 mL

of 1. 00 × 10 −2 M NaIO3 are mixed? Explain using appropriate calculations. (3 marks)

 

i think i did this one right but want to make sure.

 

A. Ksp(Cu(IO3)2)= 6.9*10^-8 Cu(IO3)2 <-  Cu2 + IO3 90.0 mL of 1. 00 10 ^2 M CuNO32.

 

10.0 mL of 1.00*10^2 M NaIO3.

 

90.0ml+10.0ml= 100(FV) Cu2(FC)=1.00*10^-2 * 90/100 = 0.9*10^-3. IO3(FC) =1.00*10^2 *

10/100 =10 tiralKsp= (0.9*10^-3) *(10)= 0.9*10^-3

 

trialKsp(0.9*10^-3) < Ksp(6.9*10^-8 )

 

a precipitate will not form

 

think i have to change thins to will form thats about it.

 

any help would be greatly appreciated.

I didn't check the math, but at a glance your methods look correct (though it's hard to read, so I could have missed something)

[boxcon]

Thx you soo much for your help I defiantly neeed to review some of the notes. I was looking over the first question and ya I could not remember why I squared that one. I realise that I should have put more sub script in to maybe make it easier to read and understand the steps that I went thought. Also maybe change the name of the post too something that would help people understand what kind of help I needed. Which would be help with equations and math not just explaining theory or fact.

can i ask for the second question if i do need to use the 250 ml sample in any of the equations to get the right Ksp for the AgCH3COO. like can i use it to get any useful information that would help me calculate the Ksp???? Or were you telling me that I should disregard it.

[hypervalent_iodine]

No, you definitely don't ignore it. I was saying that you have to use concentration of the solution, not the number of moles. You can calculate the concentration given the volume (in L) and the number of moles, which you have.

[boxcon]

o ok that makes sense sorry kind of slow sometimes but this will help a lot thank. now that im back home I can work on this a bit more

 

soo that means I should be able to get the concentration by doing this

 

1.84g*0.025L=0.046=46*10^-2g/L

 

or should it look more like this seeing as it's all ways put like this g/L

 

1.84g/0.025L=73.6g/L

 

and should I convert the grams in to moles seeing as that's just molar mass not moles MM=g??????.....nvm yes yes I should

 

 

or am I completely wrong lol

Edited June 18, 2014 by boxcon