Follow up work:
CuBr --> Cu + Br
CoBr2 --> Co + 2 Br
Set up for solubility in pure water:
5.3*10^-9 = x^2
x= 1.5 g/L
Set up (?) for solubility in 0.011 M CoBr2:
5.3*10^-9 = (x)(0.022+x)
Question: Calculate the solubility (g/L) of CuBr at 25 degrees celsius in 0.0110 M CoBr2. Ksp of CuBr = 5.3*10^-9.
I am capable of doing it in pure water, but I am struggling with how the two bromines in CoBr2 will affect the answer and ICE TABLE set up.
Thank you!
6 answers
Sorry: for solubility in water x multiplied by the molar mass equals 1.5 g/L
Your set up is right.
This is a common ion problem and the effect of the CoBr2 (Br is the common ion) DECREASES the solubility of CuBr as you would predict from LeChatelier's Principle.
CoBr2 is a soluble salt and dissociates at 100% to CoBr2 --> Co^2+ + 2Br^-
..........0.0110..0.0110....0.0220
Then for the CuBr we have
..........CuBr --> Cu^+ + Br^-
I.........solid....0......0.0220
C.........solid....x........x
E.........solid....x.....0.0220+x
Ksp CuBr = (Cu^+)(Br^-)
5.3E-9 = (x)(0.0220+x)
If you assume 0.0220 + x = 0.0220 (which is true if x is small and it is), then solve for x = (Cu^+) = (CuBr) in mols/L. Then convert to grams/L.
When you finish conmpare that to the solubility of the CuBr you worked out in pure water and see that the solubility decreased. Then look at the equation and see that you could predict that from LeChatelier's Principle.
This is a common ion problem and the effect of the CoBr2 (Br is the common ion) DECREASES the solubility of CuBr as you would predict from LeChatelier's Principle.
CoBr2 is a soluble salt and dissociates at 100% to CoBr2 --> Co^2+ + 2Br^-
..........0.0110..0.0110....0.0220
Then for the CuBr we have
..........CuBr --> Cu^+ + Br^-
I.........solid....0......0.0220
C.........solid....x........x
E.........solid....x.....0.0220+x
Ksp CuBr = (Cu^+)(Br^-)
5.3E-9 = (x)(0.0220+x)
If you assume 0.0220 + x = 0.0220 (which is true if x is small and it is), then solve for x = (Cu^+) = (CuBr) in mols/L. Then convert to grams/L.
When you finish conmpare that to the solubility of the CuBr you worked out in pure water and see that the solubility decreased. Then look at the equation and see that you could predict that from LeChatelier's Principle.
Thank so much Dr. Bob!!
X-RAY DIFFRACTION OF AN UNKNOWN METAL, MIGHT BE DANGEROUS (10/10 points)
Last week you and a friend started an experiment to obtain the X-ray diffraction peaks of an unknown metal. Through these diffraction peaks you wanted to determine:
(a) whether the cell is SC, BCC, or FCC
(b) the (hkl) value of the peaks
(c) the lattice parameter a of the metal
Unfortunately, however, your friend (since he's not in 3.091) left MIT yesterday, not to return until next semester. All of the data that you could recover from the rubble in his room was the following:
[mathjaxinline] sin^2(\theta) [/mathjaxinline] 0.120 0.239 0.480 0.600 0.721 0.841 0.956
You also know that the metal is in the cubic crystal system and the wavelength of the X-rays used is [mathjaxinline] \lambda_{CuK_{\alpha}} [/mathjaxinline]. Using the following information, determine the information you were originally interested in (a, b, and c above).
a. Is the cell SC, BCC, or FCC?
SC<text>SC</text> - correctFCCBCC
b. Enter the hkl value of the peaks as a list separated by commas. Do not put spaces between the values. For instance:
[mathjaxinline] \theta_1,\theta_2,... = (100),(111),... [/mathjaxinline]
(100),(110),(111),(200),(102),(112),(202) - correct
c. Enter the lattice parameter a in Angstroms.
2.24 - correct
Last week you and a friend started an experiment to obtain the X-ray diffraction peaks of an unknown metal. Through these diffraction peaks you wanted to determine:
(a) whether the cell is SC, BCC, or FCC
(b) the (hkl) value of the peaks
(c) the lattice parameter a of the metal
Unfortunately, however, your friend (since he's not in 3.091) left MIT yesterday, not to return until next semester. All of the data that you could recover from the rubble in his room was the following:
[mathjaxinline] sin^2(\theta) [/mathjaxinline] 0.120 0.239 0.480 0.600 0.721 0.841 0.956
You also know that the metal is in the cubic crystal system and the wavelength of the X-rays used is [mathjaxinline] \lambda_{CuK_{\alpha}} [/mathjaxinline]. Using the following information, determine the information you were originally interested in (a, b, and c above).
a. Is the cell SC, BCC, or FCC?
SC<text>SC</text> - correctFCCBCC
b. Enter the hkl value of the peaks as a list separated by commas. Do not put spaces between the values. For instance:
[mathjaxinline] \theta_1,\theta_2,... = (100),(111),... [/mathjaxinline]
(100),(110),(111),(200),(102),(112),(202) - correct
c. Enter the lattice parameter a in Angstroms.
2.24 - correct
Ma chakka muji randi ko choro ho
3 answers