June 05, 2020, 04:26:05 PM
Forum Rules: Read This Before Posting

### Topic: Can the solubility of a compound be used in conjunction with stoichiometry?  (Read 638 times)

0 Members and 1 Guest are viewing this topic.

#### Blueberries116

• Regular Member
• Posts: 30
• Mole Snacks: +0/-0
• Gender:
• Freelance scientist and learner
##### Can the solubility of a compound be used in conjunction with stoichiometry?
« on: December 30, 2019, 11:28:28 PM »
I'm confused exactly on how to use the concept of solubility to get the amount of separation from a certain solute in the solution.

This arises from attempting to solve a problem regarding this matter. The problem described is as follows:

$400$ grams of anhydrous sodium sulphate ($Na_{2}SO_{4}$) is dissolved in a liter of hot water. The solution is then let to cool carefully until reaching $20^{\circ}C$ to remain oversaturated with respect to the formed decahydrate $Na_{2}SO_{4}\cdot 10H_{2}O$. Then a small sack of the latter salt is added to the solution, separating the excess of $Na_{2}SO_{4}$ by $100\,mL$ of water. What amount of the decahydrate would had been separated?

The given alternatives are:

$\begin{array}{ll} 1.&249\,g\,\textrm{to}\,259\,g\\ 2.&318\,g\,\textrm{to}\,327\,g\\ 3.&689\,g\,\textrm{to}\,698\,g\\ 4.&721\,g\,\textrm{to}\,730\,g\\ 5.&890\,g\,\textrm{to}\,899\,g\\ \end{array}$

Although not given as information for this problem. I am assuming that the required information is the solubility of the sodium decahydrate at $20^{\circ}C$. Wikipedia entry mentions that the solubility of that salt is $44\frac{g}{100\,mL}$ for that temperature.

What I assumed is that

$400\,g\,Na_{2}SO_{4}\times\frac{142+180\,g\,Na_2SO_{4}\cdot 10 H_{2}O}{142,\g.Na_{2}SO_{4}}\,mL-100\,mL\frac{44}{100}=863\,g$

Therefore that would be the grams of sodium sulphate decahydrate but I'm not sure if that would be the ammount. I'm confused why do the alternatives features a range?. How does it appear that?. Can somebody explain this to me?. I'm not very sure about the value of the solubility of the sodium sulphate decahydrate. I've did some research on CRC's handbook and Lange's as well but there isn't given the solubility of that specific salt. Perhaps do anybody in this forum has the information, so It can match any of the alternatives.

My major concern with this question is why the answer is given in a range?. Shouldn't it be a specific amount?. What could had been intended by the author?. Can somebody help me with this?.
Have a nice day!

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7328
• Mole Snacks: +514/-86
• Gender:
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #1 on: December 31, 2019, 01:52:50 AM »
Wikipedia says that number 44g/100 ml concerns heptahydrate (but this is probably an error - should be decahydrate)

https://www.saltwiki.net/index.php/Sodium_sulfate_heptahydrate

The ranges of solubility concern eventual rounding errors.

« Last Edit: December 31, 2019, 02:09:27 AM by AWK »
AWK

#### Borek

• Mr. pH
• Deity Member
• Posts: 25780
• Mole Snacks: +1686/-400
• Gender:
• I am known to be occasionally wrong.
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #2 on: December 31, 2019, 03:01:13 AM »
Then a small sack of the latter salt is added to the solution, separating the excess of $Na_{2}SO_{4}$ by $100\,mL$ of water.

My English fails me, I have no idea what it is intended to mean
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Blueberries116

• Regular Member
• Posts: 30
• Mole Snacks: +0/-0
• Gender:
• Freelance scientist and learner
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #3 on: January 01, 2020, 04:56:45 PM »
Then a small sack of the latter salt is added to the solution, separating the excess of $Na_{2}SO_{4}$ by $100\,mL$ of water.

My English fails me, I have no idea what it is intended to mean

I'm also confused about that assertion in the question. But how could this ammount be found?. The given information on the wiki about salt is not very clear with the solubility of the sodium sulphate. I'm still stuck with this problem. Does it exist a way to solve it?.
Have a nice day!

#### Borek

• Mr. pH
• Deity Member
• Posts: 25780
• Mole Snacks: +1686/-400
• Gender:
• I am known to be occasionally wrong.
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #4 on: January 01, 2020, 05:38:43 PM »
I'm also confused about that assertion in the question. But how could this ammount be found?. The given information on the wiki about salt is not very clear with the solubility of the sodium sulphate. I'm still stuck with this problem. Does it exist a way to solve it?.

No way to say if the problem can be solved as long as we don't understand what the question really is.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7328
• Mole Snacks: +514/-86
• Gender:
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #5 on: January 01, 2020, 06:48:38 PM »
Your task is not very precise. Apart from the incomprehensible statement indicated by Borek (it is probably about washing the decahydrate with cold water - washing is not an equilibrium process and you can only calculate the maximum mass of solid salt that can be dissolved), you use the term - hot water. If you look at the Na2SO4-H2O phase diagram (https://www.saltwiki.net/index.php/Sodium_sulfate) then you can see that the minimum water temperature needed to completely dissolve anhydrous and hydrated sodium sulfate must be around 40°C - then the density of water is about 0.99 kg/L.
If you already choose the density of water, make a mass balance for it:
x_moles_Na2SO4·10·18 + (400/142-x)·1000/1.353=990 (or 1000)
(1000/1.353 is the number of grams of water per 1 mole of anhydrous sodium sulfate in a solution at 20°C - this corresponds to about 440 g of decahydrate in 1 kg of water, 1.353 is taken from the above link).
Without accurate calculation, I can estimate that the solution lies between the 2nd and 3rd alternative (with or without washing).
AWK

#### Blueberries116

• Regular Member
• Posts: 30
• Mole Snacks: +0/-0
• Gender:
• Freelance scientist and learner
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #6 on: January 01, 2020, 09:44:26 PM »
Your task is not very precise. Apart from the incomprehensible statement indicated by Borek (it is probably about washing the decahydrate with cold water - washing is not an equilibrium process and you can only calculate the maximum mass of solid salt that can be dissolved), you use the term - hot water. If you look at the Na2SO4-H2O phase diagram (https://www.saltwiki.net/index.php/Sodium_sulfate) then you can see that the minimum water temperature needed to completely dissolve anhydrous and hydrated sodium sulfate must be around 40°C - then the density of water is about 0.99 kg/L.
If you already choose the density of water, make a mass balance for it:
x_moles_Na2SO4·10·18 + (400/142-x)·1000/1.353=990 (or 1000)
(1000/1.353 is the number of grams of water per 1 mole of anhydrous sodium sulfate in a solution at 20°C - this corresponds to about 440 g of decahydrate in 1 kg of water, 1.353 is taken from the above link).
Without accurate calculation, I can estimate that the solution lies between the 2nd and 3rd alternative (with or without washing).

I'm terribly sorry for that. The question was not originally on english, so I had to translate from the original source as it was indicated. I used the word oversaturated, it should had been supersaturated.

As you indicated the statement which causes confusion is about adding the decahydrate to the solution. The amount of decahydrate is not indicated, and from looking at your solution it is not needed.

I'm looking at your mass balance equation and I still don't get it:

Can you explain it better?.

You wrote:

x_moles_Na2SO4·10·18 + (400/142-x)·1000/1.353=990

But the formatting doesn't let me to separate the intended meaning?.

Did you meant?

$$x_{moles_{Na_{2}SO_{4}}}\cdot 10 \cdot 18 + \left(\frac{400}{142-x}\right)\cdot \frac{1000}{1.353} = 990$$

I am still confused about this mass balance and its justification. I do noticed about hot water. You mentioned about minimum, how about the maximum?. The way how I'm seeing it is that it could had been let's say at 60 or 80 degrees and that salt could had been dissolved. Probably this was the intended meaning.

I've already consulted with the link you mentioned, but the only graph existing doesn't give specific values.

Regarding the $$1.35$$ per mole of anhydrous sodium sulphate:

How is it obtained: Could it be from dividing:

$$\frac{440}{232}=1.365$$

Have a nice day!

#### Blueberries116

• Regular Member
• Posts: 30
• Mole Snacks: +0/-0
• Gender:
• Freelance scientist and learner
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #7 on: January 02, 2020, 02:48:36 AM »
I'm also confused about that assertion in the question. But how could this ammount be found?. The given information on the wiki about salt is not very clear with the solubility of the sodium sulphate. I'm still stuck with this problem. Does it exist a way to solve it?.

No way to say if the problem can be solved as long as we don't understand what the question really is.

I'm terribly sorry. This problem was incorrectly stated. I had to investigate the source of that problem and found the correct version. Please take note that it was not written on english so I had to translate it to that language.

It should had been as follows:

$400$ grams of anhydrous sodium sulphate ($Na_{2}SO_{4}$) is dissolved in a liter of hot water. The solution is then let to cool carefully until reaching $20^{\circ}C$ to remain supersaturated with respect to the formed decahydrate $Na_{2}SO_{4}\cdot 10H_{2}O$. Then a small crystal of the latter salt is added to the solution, separating the excess of $Na_{2}SO_{4}$ dissolved, remaining a saturated solution. It is known that the saturated solution is equivalent to $19.4$ grams of $Na_{2}SO_{4}$ by $100\,mL$ of water. What amount of the decahydrate would had been separated?

I hope it is much clearer now. Essentially they're asking for the ammount of separation of the decahydrate. I'm still confused about how to get into the range that they propose as the answers.

with those being:

$\begin{array}{ll} 1.&249\,g\,\textrm{to}\,259\,g\\ 2.&318\,g\,\textrm{to}\,327\,g\\ 3.&689\,g\,\textrm{to}\,698\,g\\ 4.&721\,g\,\textrm{to}\,730\,g\\ 5.&890\,g\,\textrm{to}\,899\,g\\ \end{array}$

But among those four alternatives suspiciously are separated by $9\,g$ each. Could it be that there is some reason behind it?. Can you help me with this part?.
Have a nice day!

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7328
• Mole Snacks: +514/-86
• Gender:
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #8 on: January 02, 2020, 03:21:34 AM »
The solubility of mirabilite (sodium sulfate decahydrate) 1.353 mol/kg is given in Table 1 of my link.

400/142-x means (400/142)-x (see "order of operations")

In equation use: x·10·18 +.

There is no confusion concerning the seeding of a solution with crystals of decahydrate.
The confusion comes from: "separating the excess of Na2SO4 by 100 mL of water." but after you correction:19.4 is very close to 1.353·142/10=19.2 g of anhydrous sodium sulfate per 100 g of water

AWK

#### Blueberries116

• Regular Member
• Posts: 30
• Mole Snacks: +0/-0
• Gender:
• Freelance scientist and learner
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #9 on: January 23, 2020, 08:05:19 AM »
The solubility of mirabilite (sodium sulfate decahydrate) 1.353 mol/kg is given in Table 1 of my link.

400/142-x means (400/142)-x (see "order of operations")

In equation use: x·10·18 +.

There is no confusion concerning the seeding of a solution with crystals of decahydrate.
The confusion comes from: "separating the excess of Na2SO4 by 100 mL of water." but after you correction:19.4 is very close to 1.353·142/10=19.2 g of anhydrous sodium sulfate per 100 g of water

I'm revisiting this question because I'm still stuck on it  . Can you explain it in simple terms what's happening because I am still confused at the justification of your mass balance.

I could verify that from your link $19.2\,g$ of anhydrous sodium sulphate per $100\,mL$ or $100\,g$ of water is close to what was given in the problem. But the rest is where I'm still struggling. Can you please explain each term?

x_moles_Na2SO4·10·18 + (400/142-x)·1000/1.353=990

This is the equation which you used:

To which I believe the intended meaning is (after reading what you wrote):

$x\cdot 10 \cdot 18 + \left(\frac{400}{142}-x\right) \cdot \frac{1000}{1.353}=990$

Solving this produces:

$x=1.95309$

which corresponds to the moles of sodium decahydrate which translated into grams would become into.

$1.95309 \times 322 =628.498\,g$ which isn't exactly in the third option but is very close.

If I do use $1000$ instead of $990$

then this becomes into:

$1.9352 \times 322 = 623.1344\,g.$

But again, what's the justification of this equation?.

$x\cdot 10 \cdot 18 + \left(\frac{400}{142}-x\right) \cdot \frac{1000}{1.353}=990$

I'm only adding the masses of water contained in the salts?.

moles of anhydrous sodium sulphate-moles of mirabilite=moles of water

Is this what you intended to say?. Why?.

Wouldn't it be the opposite?. Why is the anhydrous first?. Or does this comes from the stoichiometry?. If it is, can you show it in the equation?.

Have a nice day!

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7328
• Mole Snacks: +514/-86
• Gender:
##### Re: Can the solubility of a compound be used in conjunction with stoichiometry?
« Reply #10 on: January 23, 2020, 08:49:17 AM »
But again, what's the justification of this equation?.

$x\cdot 10 \cdot 18 + \left(\frac{400}{142}-x\right) \cdot \frac{1000}{1.353}=990$

[Mass of water in x moles of decahydrate] + [mass of water in a remaining solution of saturated Na2SO4 at 20°C] = [mass of water used]                                      (990 g - more reliable or 1000 g - acceptable approximation)

mass of water in solution = [number of moles of anh. Na2SO4 in solution (400/142 - x)] times [mass of water per 1 mole of Na2SO4 in the saturated solution at 20°C (1000/1.353)]
« Last Edit: January 23, 2020, 09:16:30 AM by AWK »
AWK