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### Topic: Problem of the week - 01/10/2012  (Read 14865 times)

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#### Borek

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##### Problem of the week - 01/10/2012
« on: October 01, 2012, 12:08:40 PM »
Solution of a hydrohalide was neutralized by addition of the stoichiometric amount of an alkali metal hydroxide oxide. Identify the metal, if the percent concentration of the initial hydrohalide was identical to the percent concentration of the produced salt.
« Last Edit: October 08, 2012, 05:15:10 AM by Borek »
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#### Dan

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##### Re: Problem of the week - 01/10/2012
« Reply #1 on: October 02, 2012, 03:58:13 AM »
I've got it, but I don't have a very clean method yet, so I will leave it open for a bit.
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#### Borek

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##### Re: Problem of the week - 01/10/2012
« Reply #2 on: October 04, 2012, 06:38:56 AM »
Anyone else?
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#### DrCMS

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##### Re: Problem of the week - 01/10/2012
« Reply #3 on: October 04, 2012, 09:19:22 AM »
I think there are two answers depending on the strength of the HX.

If the hydrohalide solution was 25% then the metal is lithium but if the hydrohalide was 55% the metal is sodium?

#### Borek

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##### Re: Problem of the week - 01/10/2012
« Reply #4 on: October 06, 2012, 03:11:02 PM »
I got 40% hydrohalide for lithium and 71% for sodium, and as far as I can tell none of the hydrohalides can be that concentrated at STP.

Dan?

Anyone else?
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#### DrCMS

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##### Re: Problem of the week - 01/10/2012
« Reply #5 on: October 07, 2012, 04:39:39 AM »
I'll show my calculation to see if someone can see where I went wrong.

take 100g of a 25% HI solution
25g of HI is 25/(126.90447+1.00794) = 0.1954462433 moles of HI.
0.1954462433 moles of LiOH =  (0.1954462433x(6.941+15.9994+1.00794))g LiOH = 4.680613g LiOH

0.1954462433 moles of LiI = (0.1954462433x(6.941+126.90447))g LiI = 26.159594294g LiI.

26.159594294/(100+4.680613) =0.249899132 or 25%

For sodium the same calculations give 27.17%, for potassium 29.24% etc.

For 55% HI I calculate 52.18% for Li, 54.99% for Na and 57.51% for K.

at 40% HI I calculate 38.94% for Li, 41.66% for Na and 44.16% for K
at 71% HI I calculate 65.58% for Li, 68.09% for Na and 70.26% for K

I calculate it's 67.9% HX for K and 82.4% HX for Rb but I didn't think any of them other than maybe HF went that high.

#### Borek

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##### Re: Problem of the week - 01/10/2012
« Reply #6 on: October 07, 2012, 05:09:28 AM »
take 100g of a 25% HI solution

I am not questioning the approach (yet  ), I am just trying to understand why that was your starting point.
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#### DrCMS

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##### Re: Problem of the week - 01/10/2012
« Reply #7 on: October 07, 2012, 05:24:30 AM »
I just set up some calculations in a spreadsheet and using 100g of 25% solution were just the numbers I chose at random as an easy start point.

I set up the spreadsheet to use LiOH, NaOH, KOH and RbOH for HF, HCl, HBr and HI at 25% and 100g.
I can change the 25% but the 100g is fixed.
It turned out that 25% worked for Li
I then just changed the % until the Na values matched up etc etc.

I showed the 25% HI calculation as they are the bottom lines of my spreadsheet and so easy to copy out here.

#### Borek

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##### Re: Problem of the week - 01/10/2012
« Reply #8 on: October 07, 2012, 06:31:53 AM »
It is not 25 g, but 24.7744 g. Not that I understand the meaning of this number.

$$\frac{m_{LiI}}{100+m_{LiOH}}=\frac{\frac{\displaystyle m_{HI}}{\displaystyle 126.90447+1.00794}(6.941+126.90447)}{100 + \frac{\displaystyle m_{HI}}{\displaystyle 126.90447+1.00794}(6.941+15.9994+1.00794)}$$

which simplifies to

$$\frac{m_{LiI}}{100+m_{LiOH}} = \frac{5.5889247m_{HI}}{m_{HI}+534.118}$$

and as we know that

$$\frac{m_{LiI}}{100+m_{LiOH}} = \frac{m_{HI}}{100}$$

we can solve for mHI and get the value listed above.

But it disagrees with my numbers and I have no idea why.
« Last Edit: October 07, 2012, 06:44:10 AM by Borek »
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#### Dan

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##### Re: Problem of the week - 01/10/2012
« Reply #9 on: October 07, 2012, 08:31:58 AM »
I did this:

I worked in solute/solvent mass ratio:

$$\frac {m_{HX}}{m_{w1}}=\frac {m_{MX}}{m_{w1} + m_{w2}}$$

If we define the atomic mass of atom A to be RA, and moles of HX (and MX) to be n:

$$\frac {n(R_X + 1)}{m_{w1}}=\frac {n(R_M + R_X)}{m_{w1} + 18n}$$

Rearranges to:

$$\frac {n(R_X + 1)}{m_{w1}}=\frac {R_M - 1}{18}$$

or

$$\frac {m_{HX}}{m_{w1}}=\frac {R_M - 1}{18}$$

Concentration is then given by:

$$C=\frac {100(R_M - 1)}{R_M + 17}$$

So if M = Li; RM = 7; C = 25%
if M = Na; RM = 23; C = 55%
if M = K; RM = 39; C = 68%
and so on

Edit:

For more accurate values:

$$C=\frac {100(R_M - 1.00794)}{R_M + 17.00734}$$
« Last Edit: October 07, 2012, 09:51:48 AM by Dan »
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#### Borek

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##### Re: Problem of the week - 01/10/2012
« Reply #10 on: October 08, 2012, 05:13:30 AM »
It should be lithium.

Another source that I considered trusted down the drain

I have seen this question is several places. In one of them it was printed in two versions - one called for addition of Me2O for neutralization, other for addition on MeOH, both were claimed to yield the same result. Turns out MeOH version is wrong, but I have not checked it before posting - I have solved Me2O version only.

For Me2O and starting from very similar equations you both did it is possible to show that mass of HX must equal

$$m_{HX} = \frac {100} {9} (R_{Me}-1)$$

That's very similar to the latest equation listed by Dan. 100 is a mass of water (not of the solution!) in the HX solution, 9 is half of the molar mass of water, RMe is molar mass of metal. For lithium that yields around 66.6 gram of HX, and concentration of

$$\frac{66.6}{100+66.6}100\% = 40\%$$

and over 70% for sodium - so it can be safely assumed it must be Li, as 70% solutions won't be stable at STP.

Sadly, it doesn't work this way for MeOH, as you have both shown.

Edited the original wording of the problem.
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#### Rutherford

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##### Re: Problem of the week - 01/10/2012
« Reply #11 on: October 08, 2012, 07:54:48 AM »
This one reminded me of a problem I posted earlier. If someone finds now a way to solve it, please write it:
http://www.chemicalforums.com/index.php?topic=59059.0

#### Rutherford

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##### Re: Problem of the week - 01/10/2012
« Reply #12 on: October 08, 2012, 11:25:28 AM »
I don't understand the rearrangement (on the attached picture). What happens here?

#### Dan

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##### Re: Problem of the week - 01/10/2012
« Reply #13 on: October 09, 2012, 06:45:50 PM »

$$\frac {n(R_X + 1)}{m_{w1}}=\frac {n(R_M + R_X)}{m_{w1} + 18n}$$

Divide both sides by n

$$\frac {(R_X + 1)}{m_{w1}}=\frac {(R_M + R_X)}{m_{w1} + 18n}$$
$$(R_X + 1)(m_{w1} + 18n)= (R_M + R_X)(m_{w1})$$
$$R_Xm_{w1} + 18nR_X + m_{w1} + 18n = R_Mm_{w1} + R_Xm_{w1}$$
$$18nR_X + m_{w1} + 18n = R_Mm_{w1}$$
$$18nR_X + 18n = R_Mm_{w1} - m_{w1}$$
$$18n(R_X + 1) = m_{w1}(R_M - 1)$$
$$\frac {n(R_X + 1)}{m_{w1}}=\frac {R_M - 1}{18}$$
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#### Rutherford

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##### Re: Problem of the week - 01/10/2012
« Reply #14 on: October 10, 2012, 06:26:21 AM »
Thanks very much, now I am close to understand this.