How to make a tin chloride solution

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Discussion Overview

The discussion revolves around the preparation of a tin chloride solution using tin and hydrochloric acid, focusing on the proportions needed to achieve specific concentrations. Participants explore the implications of these concentrations for applications such as silvering plastic surfaces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to create a tin chloride solution without purchasing the crystal powder, asking for the correct mixing proportions of tin and hydrochloric acid to achieve a specific concentration.
  • Another participant provides an example of using 12g of pure tin with 50ml of 35% hydrochloric acid, claiming this results in a 2M solution, prompting questions about the calculation behind this concentration.
  • A participant corrects the concentration claim, stating that 12g of tin in 0.5L of acid would yield approximately 0.2M, suggesting that to achieve 2M, the tin should be dissolved in a smaller volume of acid.
  • One participant expresses the need for a lower concentration for sensitizing plastic surfaces, referencing suggestions for 0.003M and 6% solutions, and inquires about diluting the 2M solution to achieve these concentrations.
  • Participants discuss the molar mass of tin chloride (SnCl2) and its implications for calculating the mass needed for a 6% solution.
  • There is a consideration of whether the 6% concentration refers to weight by volume, and one participant suggests that for their purposes, the exact concentration may not be critical.

Areas of Agreement / Disagreement

Participants generally agree on the need for specific concentrations for different applications, but there is no consensus on the exact calculations for achieving these concentrations, particularly regarding the transition from molarity to percentage concentration.

Contextual Notes

Participants express uncertainty about the exact proportions and calculations needed to convert between molarity and percentage concentration, particularly in the context of preparing solutions for specific applications.

lovicodo
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hello, i would like to make a tin chloride solution without having to buy the tin chloride crystal powder.
i know that tin and hcl are enough, but i would like to know how they can mixed ,and in which proportion, to get a specific concentration.

i found an example where are used:
12gr pure tin
50cl hydrochloric acid 35% solution
the author claims that in this way the final concentration will be 2M
someone can explain me why the final concentration will be 2M with this proportion of tin and hcl?
thank you!
 
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lovicodo said:
hello, i would like to make a tin chloride solution without having to buy the tin chloride crystal powder.
i know that tin and hcl are enough, but i would like to know how they can mixed ,and in which proportion, to get a specific concentration.

i found an example where are used:
12gr pure tin
50cl hydrochloric acid 35% solution
the author claims that in this way the final concentration will be 2M
someone can explain me why the final concentration will be 2M with this proportion of tin and hcl?
thank you!
What is your background in handling chemicals? Have you read the MSDS for tin chloride and the components you want to use to synthesize it? What kind of a lab will you be using, and what PPE will you be using?

https://www.sigmaaldrich.com/catalog/product/sial/452335?lang=en&region=US

MSDS: https://www.sigmaaldrich.com/MSDS/M...drich.com/catalog/product/sial/452335?lang=en

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i need this solution to try to do silvering of a plastic surface for my hobby project
the tin solution in this case will be used as 'sensitizer', i.e. it willbe used to prepare the surface to be silvered...
i have only a scholastic background, not a professional/expert background
thank you for the link and the safety sheet
 
lovicodo said:
hello, i would like to make a tin chloride solution without having to buy the tin chloride crystal powder.
i know that tin and hcl are enough, but i would like to know how they can mixed ,and in which proportion, to get a specific concentration.

i found an example where are used:
12gr pure tin
50cl hydrochloric acid 35% solution
the author claims that in this way the final concentration will be 2M
someone can explain me why the final concentration will be 2M with this proportion of tin and hcl?
thank you!
Well, tin has a molecular mass of 119 g/mol. So 12 g are about 1/10th of a mole. Dissolved in 0.5 l of acid makes a solution with about 0.2 M (=mol/L) and not 2 M. So you should dissolve the tin in 50 ml (or even a bit less) to get a 2M solution.
The solution will be rather harmless unless you drink it completely. Nevertheless you should use safety googles especially when handling the acid.
 
Thank you very much for the answer
yes, you are right, i did a mistake, i wrote 50cl, but it was 50ml.
About my project:
2M solution is very concentrated, this is needed only to melt tin.
To sensitize the plastic i need lower concentrations:
- an author suggests to make a 0.003M solution and keep the plastic immersed in the solution for about 30 minutes
- an other author suggests to make a 6% solution and keep the plastic immersed for no more than 30 seconds
so i need to take only a small portion of the 2M solution and add the right quantity of water :
for example, in the first case (0.003M), i could pick 1ml of 2M , and add (2/0.003 - 1) ml = 665 ml water, i am right?
but in the second case? 6% thin chloride solution what does mean?
 
it is 6% in weight?
how can i get a 6% tin chloride solution starting from a 2M solution?
thanks
 
What is the molar mass of SnCl2?
 
189,6 g/mol
so i have about 380g SnCl2 in 1L of the 2M solution...
 
lovicodo said:
189,6 g/mol
so i have about 380g SnCl2 in 1L of the 2M solution...
Very good. So if you want 6 % by mass?
 
  • #10
this is the question :)
perhaps to know the concentration in percent i need to know the mass of 1L of the 2M solution...
i can guess...380g is 6% of (380/0.06)g = 6333g so i have to add 6333 - 380 = 5953g water, that is about 6L water.
So total volume is 6Lwater+380g tin chloride..i don't know the exact result , in volume, of this sum...
Very roughly, assuming volume is about the same (6L), i guess that i have to add about 5L water to 1L 2M solution to get a 6% tin chloride solution...
that is, add 5 parts of water for each part of 2M solution...

is it right?
but if i would like to know the exact proportions, how could i calculate them?

thanks!
 
Last edited:
  • #11
I just saw that SnCl2 solution is commercialized as weight per volume
http://www.labchem.com/tools/msds/msds/LC25180.pdf
If you assume that these 6% are weight by volume, you don't need to know the density of the final solution.
For your purposes (pre treatment of surfaces to be coated by silver) the exact concentration will be irrelevant, anyhow.
 
  • #12
thank you,
i think so too
 

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