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Anna [14]
3 years ago
7

What reagent separates Ag+ from Cu+ and Cu2+ from Bi3+? Explain the chemistry.

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
5 0
 <span>to separate Ag+ from Cu2+___________________________ 

treat the solution of Ag+ and Cu2+ with cold dilute HCl 

Ag2+ will get precipitated as AgCl (white colour precipitate) 
[Ksp of 1st group cation chlorides are low] 

filter the solution and remove the precipitate 

then pass H2S through the already acidified solution 

Cu2+ will get precipitated as CuS (Black color precipitate) 
[Ksp of 2nd group cation sulphides are low] 


to separate Cu2+ from Bi3+______________________________ 

add a few drops of dilute HCl into the solution of Cu2+ and Bi3+ 

BiCl3 is formed ( CuCl doesnt form because because Ksp value of BiCl3 is less than Ksp value of CuCl) 

then dilute the solution by adding water 

BiCl3 react with water and forms BiOCl and get precipitated (white color precipitate) 

filter the solution and remove the precipitate 

then pass H2S through the already acidified solution 


Cu2+ will get precipitated as CuS (Black color precipitate) 
[Ksp of 2nd group cation sulfides are low]</span>
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In this case we can start with the <u>ppm formula</u>:

ppm=\frac{mg~of~solute}{Litters~of~solution}

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of F^-, because the molarity formula is:

M=\frac{mol}{L}

In other words:

0.0320~M=\frac{mol}{1~L}

mol=0.032~M*1~L=0.032~mol

1~L~of~Solution=0.0320~mol~of~solute

If we use the <u>atomic mass</u> of F  (19 g/mol) we can convert from mol to g:

0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

0.607~g\frac{1000~mg}{1~g}=607~mg

Finally we can <u>divide by 1 L</u> to find the ppm:

ppm=\frac{607~mg}{1~L}=~607~ppm

<u>We will have a concentration of 607 ppm.</u>

I hope it helps!

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