Respuesta :
Ksp of SrSO4 = 3.2 x 10-7, so it will precipitate when Ksp > 3.2 x 10-7
Same way, calculating for 0.0150 M Sr2 SrSO4 precipitates,
when [ SO4 2-] > (3.2 x 10-7 / 0.0150) = 2.133 x 10^-05.
Thus SrSO4 precipitates when SO4 2- > 2.133 x 10^-05
Now the Pb2+ at this point = 2.133 x 10^-05 x 0.0150 =3.2 x 10^-07
Hence Ksp for PbSO4 = 1.6 x 10-8
the solubility product for PbSO4 is more and the PbSO4 precipitates as
follows [Pb2+] = 1.6 x 10-8 / 2.133 x 10^-05 = 7.50 x 10^-04
Same way, calculating for 0.0150 M Sr2 SrSO4 precipitates,
when [ SO4 2-] > (3.2 x 10-7 / 0.0150) = 2.133 x 10^-05.
Thus SrSO4 precipitates when SO4 2- > 2.133 x 10^-05
Now the Pb2+ at this point = 2.133 x 10^-05 x 0.0150 =3.2 x 10^-07
Hence Ksp for PbSO4 = 1.6 x 10-8
the solubility product for PbSO4 is more and the PbSO4 precipitates as
follows [Pb2+] = 1.6 x 10-8 / 2.133 x 10^-05 = 7.50 x 10^-04
The precipitate is an insoluble solid formed in a supersaturated solution. The concentration of lead (II) ion when strontium sulfate begins precipitating is [tex]7.50 \times 10^ {-4}.[/tex]
What is Ksp?
Ksp is the solubility product constant of solid solute in an aqueous medium. It depicts the solubility of the substance in the aqueous solution.
We know that the solubility constant of strontium sulfate = [tex]3.2 \times 10^{-7}.[/tex]
Calculating the solubility constant of 0.0150 M strontium sulfate:
[tex][ \rm SO_{4} ^{2-}] > (\dfrac{3.2 \times 10^{-7}}{0.0150}) = 2.133 \times 10^{-5}[/tex]
Hence, strontium sulfate will precipitate when [tex][ \rm SO_{4} ^{2-}] > 2.133 \times 10^{-5}[/tex]
Now, Ksp of lead will be:
[tex]2.133 \times 10^{-5} \times 0.0150 =3.2 \times 10^{-7}[/tex]
The solubility product of Lead sulfate is greater and will precipitate when,
[tex]\dfrac{1.6 \times 10^{-8} }{2.133 \times 10^{-5}} = 7.50 \times 10^{-4}[/tex]
Therefore, the lead will precipitate at a concentration of [tex]7.50 \times 10^ {-4}.[/tex]
Learn more about precipitation here:
https://brainly.com/question/14152510
