Wednesday

October 7, 2015
Posted by **Kizner** on Sunday, June 11, 2006 at 3:18pm.

Below, I write M.S. (molecular solubility)

1)Sr3(AsO4)2<---> 3Sr^2+ + 2AsO4^-3

M.S. xmol/L 3x 2x

(3x)^3 (2x)^2 = 108x^5

(kindly explain why it is (3x^3)(2x)^2 as I followed an diff. example from the book)

Solving for x I get 1.04E-4

So the molar solubility: 1.04E-4 mol Sr3(AsO4)2/L (dissolved); then the concentrations of the ions are:

[Sr^2+] = 3x = 3.1E-4

[AsO4^3-] = 2x = 2.1E-4

If these are correct than your help is starting to sink in, if not than a little more practice should do. By the way, I don't mind lenghty detailed replies, it only shows that you are as passionate about the subject as I'd like to be. You must be a chemist! Thx again Dr.Bob Also, I imagine that anyone is free to reply to a post; I have replied to some chemistry and math posts, but I am not sure if I'm breaking any rules. It is a fun website and I do learn. Six week course, particularly chem II is rough. Way too fast. :)We're coverring

"Eclectro chemistry, metals II and coordination compounds". That make three chapters in two more days before dreaded finals next monday! Phew! Wish me luck!

Your work is impeccable. I THINK you will understand the (3X)^3(2x)^2 business if you had written the Ksp expression. For this reaction it is

Ksp = (Sr^+2)^3(AsO4^-3)^2.

(Sr^+2) = 3X and that is cubed in the Ksp expression so it becomes (3X)^3.

(AsO4^-3) = 2X and that is squared in the Ksp expression to become (2X)^2

and voila!, (3X)^3(2X)^2 = etc.

This will get your quick answer out and I will address the remainder of your questions in another post.

Yes, anyone is free to post or to respond to a post. You aren't breaking any rules to respond to a post. Of course, posting nonsense replies or smart alecky replies is not allowed and those replies will be removed. On another matter, often you can find a problem posted here that is similar to your own and you learn by reading the answers to those posts. Finally, yes, I am proud to say I am a chemist and have been for many more years than your age. I have learned some things through experience and many things through formal education and a few more things through research. I also learned quite a bit of chemistry through teaching. More than once I have been explaining something to a class about some topic when the light dawns about a similar subject, and, without any change in my voice or pace or speaking, I think "Oh, so that's why that(fill in the blank ________ works). From then on those two topics are then connected irreversibly in my mind and I have the luxury of understanding a little more about what I am talking about. I have been retired 11 years from the university where I taught and I like to do this to help those who truly want to be helped and aren't just looking for a free hand at slapping some answers down on a complete the worksheet type homework assignment. All of us who work this site are volunteers and I suspect all are former/current teachers. Thank you for using this web site and please know that you and others are most welcome to post your questions here. You get the most attention when posting a question to do as you have done and show us your work. We can evaluate a question much more completely if we know what you are thinking. Good luck on your exams. That's always a traumatic time but the secret is to be prepared.

- Dr.Bob (thx) Check my work for this one pls -
**Allison**, Tuesday, October 30, 2007 at 7:14pmOh my god i am not in 8th grade

- Dr.Bob (thx) Check my work for this one pls -
**Allison**, Tuesday, October 30, 2007 at 7:15pmI love you

- Dr.Bob (thx) Check my work for this one pls -