Respuesta :
Answer:
[tex]Molar solubility(S)=1.71\times10^{-2} mole/litre[/tex]
Explanation:
Assume Ksp of [tex]Pb(SCN)_{2}[/tex]=[tex]2\times10^{-5}[/tex]( it is general value ,i took it because not given in question)
molar solubilty means how much maximum mole of lead thiocynate is dissolved in one litre of solvent.
Ksp=solubility product;
[tex]Pb(SCN)_2= Pb^{2+} +2SCN^-[/tex]
Lets molar solubilty of [tex]Pb(SCN)_2[/tex] is S then
[tex][Pb^{+2}]=S;[/tex]
[tex][SCN^{-1} ]=2S;[/tex]
[tex]Ksp =S\times(2S)^{2} =2\times10^{-5}[/tex]
[tex]4S^{3}=2\times10^{-5}[/tex]
[tex]S=1.71\times10^{-2} mole/litre[/tex]
The molar solubility of lead thiocyanate in pure water is 1.71*10^-2 mole/litre.
Calculation of the molar solubility:
Here we assume
K-s-p of P-b-(S-C-N) = 2*10^-5
So,
P-b(S-C-N)_2 = P-b^2 + 2S-C-N^-
Here we assume molar solubility should be S
So,
{Pb^+2} = S
S-C-N^-1 = 2S
k-s-p = S*(2S)^2
= 2*10^-5
4S^3 = 2*10^-5
So, S = 1.71*10^-2
hence, The molar solubility of lead thiocyanate in pure water is 1.71*10^-2 mole/litre.
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