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Please check to see if there is anything that I should fix. Also, please check to make sure that my significant figures are correct. I am going to try to make it look like how I wrote it and how the book shows the information. Please don't fix what I wrote I am trying to show you how I wrote it out and I don't want you to be confused.

Please use this link if you want a better look at the equations and parts.

https://books.google.com/books?id=X3KX6p8hnlIC&pg=PA304&lpg=PA304&dq=Five+repeated+calculations+are+required+to+establish+the+data+plot+for+the+determination+of&source=bl&ots=AenLh4IGsI&sig=hs7pdP6AnhvgaTVs0zvcGChbTzE&hl=en&sa=X&ved=0ahUKEwjzpZ_WlKrJAhWBHD4KHQ-ICqIQ6AEIHDAA#v=onepage&q=Preparation%20of%20Borax%20solutions&f=false

For question 3. a., it says "See equation 11.11". In the link it is on page 301 and called "(26.11)" instead of being called "(11.11)".

For question 3. b., it says "See equation 11.12". In the link it is on page 301 and called "(26.12)" instead of being called "(11.12)".

For question 3. c., it says "See equation 11.15". In the link it is on page 301 and called "(26.15)" instead of being called "(11.15)".

For question 3. d., it says "See equation 11.4". In the link it is on page 300 and called "(26.4)" instead of being called "(11.4)".

  1. The $\ce{B4O5(OH)_4^2-}$ ion, present in $5.0\ \mathrm{mL}$ of a saturated $\ce{Na2B4O5(OH)4}$ solution at a measured temperature, is titrated to the bromocresol green endpoint with $5.28\ \mathrm{mL}$ of $0.182\ \mathrm{M}\ \ce{HCl}$. Express all calculations with the correct number of significant figures.

a. How many moles of $\ce{B4O5(OH)_4^2-}$ are present in the sample? See equation 11.11.

My Work:

(In case you don't look at the link for equation 26.11 which is the same as 11.11 this is kind of what is should look like on here just to show it a bit) mol $\ce{B4O5(OH)_4^2-}$ = volume (L)HCl X mol HCl X 1 mol $\ce{B4O5(OH)_4^2-}$

(What I am typing now please think about this. I wanted to but didn't know how to write this out onto the above equation. Put a volume(L)HCl (this is another volume(L)HCl being used in the equation don't take the one from the beginning and put it underneath the mol HCl) under the mol HCl like it is dividing and put a 2 mol HCl under the 1 mol $\ce{B4O5(OH)_4^2-}$ like it is dividing. That is how the above equation should look like once put together.)

mol $\ce{B4O5(OH)_4^2-}$ = 0.00528L HCl X 0.00096096mol HCl X 1 mol $\ce{B4O5(OH)_4^2-}$

(What I am typing now please think about this. I wanted to but didn't know how to write this out onto the above equation. Put 0.00528L HCl under the 0.00096096mol HCl like it is dividing and then put 2 mol HCl under the 1 mol $\ce{B4O5(OH)_4^2-}$ like it is dividing. That is how the above equation should look like once put together.)

= 0.00048 moles of $\ce{B4O5(OH)_4^2-}$ are present in the sample

The number of moles of $\ce{B4O5(OH)_4^2-}$ that are present in the sample are 0.00048 moles of $\ce{B4O5(OH)_4^2-}$ are present in the sample.

b. What is the molar concentration of $\ce{B4O5(OH)_4^2-}$ in the sample? See equation 11.12.

My Work:

(In case you don't look at the link for equation 26.12 which is the same as 11.12 this is kind of what is should look like on here just to show it a bit)

[B4O5(OH)42-] (this should look like the equation below with the brackets around it) = mol $\ce{B4O5(OH)_4^2-}$ / volume(L) sample

In the above equation I want to clarify everything. The "[B4O5(OH)42-]" should look like the bracketed one below. Then "=" "mol $\ce{B4O5(OH)_4^2-}$" and underneath the "mol $\ce{B4O5(OH)_4^2-}$" is volume(L) sample.

$$\ce{[B4O5(OH)_4^{2-}]} = \frac{0.00048\ \mathrm{mol}\ \ce{B4O5(OH)_4^2-}}{0.005\ \mathrm{L sample}}$$

= 0.096 molar concentration of $\ce{B4O5(OH)_4^2-}$ in the sample

The molar concentration of $\ce{B4O5(OH)_4^2-}$ in the sample is 0.096 molar concentration of $\ce{B4O5(OH)_4^2-}$ in the sample.

c. Calculate the $\mathrm{K_{sp}}$ for $\ce{Na2B4O5(OH)4}$ from these data. See equation 11.15.

My Work:

(I don't know which way is right so I wrote "OR")

(In case you don't look at the link for equation 26.15 which is the same as 11.15 this is kind of what is should look like on here just to show it a bit)

Ksp = [Na+][B4O5(OH)42-] = [2 X [B4O5(OH42-]]^2[B4O5(OH)42-] and continued is here okay = 4[B4O5(OH)42-]^3 = 4[molar solubility of borax]^3

(0.00048 + 0.096) X 4

= 0.38592

OR

4 X 0.00048^3

= 4.42368 E-10

d. What is the free energy change for the dissolution of $\ce{Na2B4O5(OH)4}$ at $25\ \mathrm{C}^\circ$? See equation 11.4. R = 8.314 X 10^-3 kJ/mol•K.

My Work:

(In case you don't look at the link for equation 26.11 which is the same as 11.11 this is kind of what is should look like on here just to show it a bit)

Δ°G = -RTlnKsp = -RTln[Ag^+]^2[CrO42-]

Δ°G = -(0.008314kJ•mol^-1^-1)(298K)(0.38592)

= -0.9561445862

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  • $\begingroup$ Welcome to Chemistry Stack Exchange. I have edited your question so that it is in the correct format. I have also removed the links and information regarding the equations as this being a chemistry website, I sure that most people are aware of the equations already. However if you feel that they are necessary, feel free to add them again. $\endgroup$ – Nanoputian Nov 25 '15 at 20:49
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    $\begingroup$ No worries. Your answers for (a) and (b) are correct, but the answer in part (a) should be to 3 sig figs. So the answer should be $4.80\times 10^{-4}$. Also you forgot to put units for your answer which is extremely important (more important than correct sig figs). $\endgroup$ – Nanoputian Nov 25 '15 at 20:53
  • $\begingroup$ Unfortunately I probably won't be able to write a complete answer for parts (c) and (d) right now as I have to go to school now. In the afternoon, if no one has answered this question, I will write one up. $\endgroup$ – Nanoputian Nov 25 '15 at 20:54
  • $\begingroup$ Thank you. I did these and just wanted to make sure that I wasn't messing up anywhere. $\endgroup$ – Josh Palmer Nov 25 '15 at 20:55
  • $\begingroup$ Please look and see if there is anything else that needs to be fixed. $\endgroup$ – Josh Palmer Nov 27 '15 at 16:53
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I like Nanoputian's answers. But I think he/she may have made a mistake for part C. I got 3.5E-3 (instead of 3.5E-2). When it comes to part D. I noticed your answer was crazy wrong. Here's a qualitative check for this problem. You know spontaneous processes have negative changes in Gibbs Free Energy and the converse is true for non-spontaneous processes. Notice that the concentration of borax in solution was low, which suggest the process isn't spontaneous. If the process was spontaneous, dissolution wold be facile and most of the borax would be in solution. If you'd like a more quantitative check. Find the case were ln(K) = 0, or when K = 1. If 4x^3 = 1 then x = 0.63. The concentration of borax should be 0.63 M when ln(K) = (delta)G = 0. A borax concentration less than the previously specified concentration yields a positive change in Free Energy. The converse would be true if the concentration was higher. I hope this info helps on future problems like this.

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  • $\begingroup$ Thanks for the correction Joel! Thats what happens when you are rushing to write an answer :) $\endgroup$ – Nanoputian Nov 30 '15 at 9:39
  • $\begingroup$ However I fail to see where I went wrong for part d. Could you please explain what I did wrong ? $\endgroup$ – Nanoputian Nov 30 '15 at 9:45
  • $\begingroup$ Is my part d correct now? Or at least one of the answers I provided. $\endgroup$ – Josh Palmer Dec 1 '15 at 1:01
  • $\begingroup$ I'm not entirely sure how you got ln(K) = 0.38592 because that implies K = 1.47. My suggestion is that you make sure the value for K is correct and that you use the natural logarithm function in your calculator. $\endgroup$ – Joel J. Dec 1 '15 at 2:24
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Before I start going through each question, let me stress the importance of units in your answer and in your working out. It is good to see that you are putting in the units in your working out (this is something that even I fail to do), however it is also very important to include the correct units in your answer as you will definitely loss marks if you don't.

Also generally it is better to write you answer using scientific notation, especially for really small numbers. This is because it much clearer and it is also easier to show the correct number of sig figs.

Question (a)

I don't really understand the equation that they have given you since they merge several steps to form one big equation. You shouldn't rely on formulas as you won't be able to answer other questions which might require a slightly different method. I will show you how I would do this question, step by step.

First of all, write down the chemical equation:

$$\ce{B4O5(OH)_4^{-2} + 2H+ -> H2B4O5(OH)_4}$$

By looking at the stoichiometry, for every mole of $\ce{B4O5(OH)_4^{-2}}$ present, $2$ moles of $\ce{H+}$ is required. So now lets calculate the amount of moles of $\ce{H+}$:

$$\mathrm{n}(\ce{H+}) = 5.28\times 10^{-3}\ \mathrm{L} \times 0.182\ \mathrm{mol\ L^{-1}} = 9.61 \times 10^{-3}\ \mathrm{mol}$$

Since we know that the amount of moles of $\ce{B4O5(OH)_4^{-2}}$ is half the amount of moles of $\ce{H+}$:

$$\mathrm{n}(\ce{B4O5(OH)_4^{-2}}) = \frac{1}{2} \times 9.61 \times 10^{-3}\ \mathrm{mol} = 4.80\times 10^{-4}\ \mathrm{mol}$$

Note: that the correct answer should be to 3 sig figs, not 2 since all the data we used in this question is to 3 sig figs.

Question (b)

There is really nothing to say for this question. Just remember that your answer should be $9.6\times 10^{-2}\ \mathrm{M}$.

Question (c)

For calculating $\mathrm{K_{sp}}$, use the second formula that you used. However you need to put in the value of the concentration of $\ce{B4O5(OH)_4^{-2}}$, not the amount of moles. So the correct answer should be: $3.5\times 10^{-3}$

Note: there is no units for $\mathrm{K_{sp}}$ since it is really just a ratio.

Question (d)

The formula for free energy change is: $$\Delta G = -\mathrm{R}T\mathrm{ln}K$$ However when subbing the values in the equation, you forgot to use $\mathrm{ln}K$. Therefore the answer should be: $14\ \mathrm{kJ}$

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  • $\begingroup$ Let me fix what some other people edited in my post. It doesn't look like how it should. Some things are there now that aren't in my book. $\endgroup$ – Josh Palmer Nov 26 '15 at 14:54
  • $\begingroup$ Please look and see if there is anything else that needs to be fixed now that I made it look like my work and the book. $\endgroup$ – Josh Palmer Nov 27 '15 at 16:53
  • $\begingroup$ I think part A you went to say 4.80×10−4 mol. $\endgroup$ – Josh Palmer Nov 29 '15 at 22:45

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