Lead metal is added to 0.140 Cr^3+(aq).
Pb(s)+ 2Cr^3(aq)<---> Pb^2+(aq)+2Cr^2+(aq)
KC=3.2*10^{-10}
A. What is Pb^2+ when equilibrium is established in the reaction?
B. What is Cr^2+ when equilibrium is established in the reaction?
C. What is Cr^3+ when equilibrium is established in the reaction?
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Chemistry please help - DrBob222, Friday, February 6, 2009 at 7:43pm
Do an ICE chart.
Initial:
Cr^+3 = 0.140
Cr+2 = 0
Pb^+2 = 0
Change:
Cr^+3 = -2y
Cr^+2 = +2y
Pb^+2 = +y
Equilibrium:
Cr^+2 = +2y
Pb^+2 = +y
Cr^+3 = 0.140-2y
Plug all of that into the expression for Kc and solve for y. I think it's a cubic equation.
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SO this is my ice chart:
2Cr^3+ 2Cr^2+ Pb^2+
Initial 0.140 0 0
Change -2y +2y +y
Equlibrium 0.140-2y +2y +y
Kc= {Pb^2+}{cr^2+}^2/ {Cr^3+}^2
so 3.2*10^-10= (y)(2y)^2/(0.140-2y)^2
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When i solve this im not getting the right answer i posted it up as well,
and these are the answers i got, but these are wrong as well..if you could please help:
Answer #1:
3.2*10^-10= (y)(2y)^2/ (0.140-2y)^2 or
3.2*10^-10= 2y^3/ (0.140-2y)^2
for all practical purposes, the left side of your equation is "zero"
(it has value 0.00000000032)
so we end up with 2y^3 = appr. 0
so y = appr. 0
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Answer # 2
(0.140-2y)^2 * (3.2*10^-10) = 4 y^5
(0.0196 -0.280y + 4y^2)*(3.2*10^-10)
= 4 y^5
6.282*10^-12 -8.96*10^-11 y +1.28*10^-9 y^2
= 4 y^5
This is a messy fifth order polynomial that is best solved by iteration. As a first approximation, I would assume 4 y^5 = 6.282*10^-12, since the y and Y^2 terms will be negligible compared to the constant.
y^5 = 1.57*10^-12
y = 4.36*10^-3
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And i have a different question regarding a post someone else put up yesterday:
________________
Hey guys I did a lab and I have to answer the questions relating to the lab. I did question 1 & 2, can someone check if they are correct, and question 3 I don't know how to do it.
1. Calculation to Determine the molecular weight of unknown substance
Mass of unknown used: 2.0 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -1°C
Solution = delta Tf/-Kf
= -1°C/-1.86°C kg/mole
= 0.538 m (molality)
Molar mass = 2g / 0.05 kg * 0.538 m
= 74.4 g/mol
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These are my values:
Mass of unknown used:1.9397 g
Mass of water used: 50.0 g (0.05kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -.10°C
I did this lab as well, So i was wondering when it says:
Solution = delta Tf/-Kf
Wouldn't delta be final temp - initial temp:
So my initial temp was 23.00C and final -.10:
So : is delta T = -23.1
Solution = delta Tf/-Kf
= -23.1°C/-1.86°C kg/mole
= 12.4 m (molality)
Molar mass = 1.9397g / 0.05 kg * 12.4 m
= 481.04 g/mol
Which is totally wrong so now i m confused
With regard to question #2.
Yes, this was a previous post which I didn't answer but Bob Pursley did. He suggested that the student didn't have good precision in the measurements. In answer to your question, yes, delta T is 23.1 if you started with 23.00 and ended up with -0.1 (but it's +23.1 if that makes a difference because 23.00-(-0.1) = 23.00 + 0.1 = 23.1 degrees T. Next, if I'm able to separate what the other post was from your data, then I don't get your answer of 481.04 but my answer is ridiculous.
I agree with 12.4 m
Then m = mols/kg = 12.4 = mols/0.05
Solve for mols = 12.4 x 0.05 = 0.62
mols = g/molar mass = 0.62 = 1.9397/molar mass
Solve for molar mass = 1.9397/0.62 = 3 something and you know that can't be right. I don't remember the details of the problem but one possible problem could be that ionic compounds have an i in the equation (for number of particles). The second problem is simply that it's a lab procedure and we have no way of knowing how well the numbers have been measured. One comment I have, however, is that 12.4 m seems to be a high value for a 2 g sample in 50 mL water. But then I'm not familiar with the experiment and how it was conducted.
For your number 1 problem, the first thing I see is that you have y(2y)^2 = 2y^3 but it should be 4y^3. Can you repost the problem so I can look at all of the details again. I've forgotten the details. One of the answers you had returned was that with the left side being so small that y would be essential zero BUT if you look at Kc, with a value like 10^-8, the products WILL be small and could well approach zero.
This is the original post:
Posted by Saira on Friday, February 6, 2009 at 6:11pm.
Lead metal is added to 0.140 Cr^3+(aq).
Pb(s)+ 2Cr^3(aq)<---> Pb^2+(aq)+2Cr^2+(aq)
KC=3.2*10^{-10}
A. What is Pb^2+ when equilibrium is established in the reaction?
B. What is Cr^2+ when equilibrium is established in the reaction?
C. What is Cr^3+ when equilibrium is established in the reaction?
Responses
* Chemistry please help - Saira, Friday, February 6, 2009 at 6:13pm
* correction 0.140 M Cr^3+(aq).
* Chemistry please help - DrBob222, Friday, February 6, 2009 at 7:43pm
You didn't make a correction with your correction.
Pb(s) + 2Cr^+3 ==> 2Cr^+2 + Pb^+2
Do an ICE chart.
Initial:
Cr^+3 = 0.140
Cr+2 = 0
Pb^+2 = 0
Change:
Cr^+3 = -2y
Cr^+2 = +2y
Pb^+2 = +y
Equilibrium:
Cr^+2 = +2y
Pb^+2 = +y
Cr^+3 = 0.140-2y
Plug all of that into the expression for Kc and solve for y. I think it's a cubic equation.
OK.
I stayed with my original equation which is
Kc = y*(2y)^2/(0.140-2y)^2 = 3.2 x 10^-10.
Of course we agree that the 3.2 x 10^-10 is a very small number and that times the denominator term = essentially zero which means y is approximately zero for y, BUT (and this is a big BUT) what's so bad about that? With a Kc of 3.2 x 10^-10 it means the equilibrium is FAR FAR to the left which means not much of the products are formed so a number approximately zero would be expected. Isn't that correct? At any rate, I solved the cubic equation. Did you expand the denominator, collect terms, and solve the cubic or did you just post the expression which led to the comments about the "left side being essentially zero etc etc"? Anyway, expand the denominator. You do it. You need to cherck my work ANYWAY. Here is what I obtained.
4y^3 -1.28 x 10^-9(y^2) + 1.792 x 10^-10(y) -6.272 x 10^-12 = 0
Now solve for y and I have 0.000116.
I plugged that number back into the cubic and it satisfies the cubic equation that we generated. That only proves that the value is the correct solution of the equation; it doesn't say our chemistry is right about how we obtained the cubic in the first place. Then I determined 2y (for Cr^+2) and (0.140-2y) for (Cr^+3), plugged those into the Kc expression and I obtained 3.2 x 10^-10 WHICH IS Kc. So if 0.000116 gets Kc, that's pretty convincing evidence that the chemistry is correct. One note here is that when working with such small numbers, you can't just throw away those pesky numbers far to the right of the decimal point AND when subtracting small numbers from the 0.140, you MUST carry it out to more places than allowed (significant digit wise) because we know that leaving 0.140 as that number we can't possibly get Kc. Anyway, check my work. Check for typos and check for math errors. But I think this is right.
For the question about the equilibrium of the reaction between lead metal and chromium ions, let's go through the solution step by step:
1. Start with the reaction equation:
Pb(s) + 2Cr^3+(aq) ⇌ Pb^2+(aq) + 2Cr^2+(aq)
2. Use an ICE chart to set up the initial, change, and equilibrium values for the concentrations:
Initial:
Cr^3+ = 0.140 M
Cr^2+ = 0 M
Pb^2+ = 0 M
Change:
Cr^3+ = -2y
Cr^2+ = +2y
Pb^2+ = +y
Equilibrium:
Cr^3+ = 0.140 - 2y
Cr^2+ = 2y
Pb^2+ = y
3. Write the expression for the equilibrium constant, Kc:
Kc = ([Pb^2+]*[Cr^2+]^2) / ([Cr^3+]^2)
4. Substitute the equilibrium concentrations from the ICE chart into the equilibrium constant expression:
Kc = (y*(2y)^2) / (0.140-2y)^2
To solve for the value of y (concentration of Pb^2+), we would need to solve this equation for y. However, it looks like there might be a mistake in the equation setup you provided. The correct equation should be:
Kc = (4y^3) / (0.140-2y)^2
You can solve this cubic equation for y using numerical methods or approximation techniques. There doesn't seem to be an analytical solution in this case.
Regarding your second question about calculating the molecular weight of an unknown substance, let's go through the steps:
1. Given values:
Mass of unknown used: 1.9397 g
Mass of water used: 50.0 g (0.05 kg)
Kf = 1.86°C kg/mole
Experimentally determined freezing point of unknown: -0.10°C
(Note: I will use your given mass of the unknown substance for calculation.)
2. Calculate the change in temperature (delta T):
delta T = final temperature - initial temperature
delta T = -0.10°C - 23.00°C
delta T = -23.1°C
3. Calculate the molality (m) of the solution:
molality (m) = delta T / -Kf
molality (m) = -23.1°C / -1.86°C kg/mole
molality (m) = 12.4 mol/kg
4. Calculate the molar mass of the unknown substance:
Molar mass = mass of unknown substance / mass of water * molality
Molar mass = 1.9397 g / 0.05 kg * 12.4 mol/kg
Molar mass = 481.04 g/mol
It looks like your calculations for the molar mass are correct. The discrepancy in the answer might be due to rounding errors or differences in the given values. Make sure to double-check the given values and perform the calculations accurately.