Respuesta :
Answer:
[tex][Ag^{+}]=4.2\times 10^{-2}M[/tex]
Explanation:
Given:
[AgNO3] = 0.20 M
Ba(NO3)2 = 0.20 M
[K2CrO4] = 0.10 M
Ksp of Ag2CrO4 = 1.1 x 10^-12
Ksp of BaCrO4 = 1.1 x 10^-10
[tex]BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)[/tex]
[tex]Ksp=[Ba^{2+}][CrO_{4}^{2-}][/tex]
[tex]1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}][/tex]
[tex][CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}[/tex]
Now,
[tex]Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)[/tex]
[tex]Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}][/tex]
[tex]1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})[/tex]
[tex][Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}[/tex]
[tex][Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M[/tex]
So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M
The concentration of silver ions in the solution when the Barium chromate starts to precipitate is 0.042 M.
What is solubility equilibrium?
The solubility equilibrium has been the concentration of the ions at which the rate of solubility and the precipitation are equivalent.
The concentration of chromate ion with barium ion concentration of 0.2 M is:
[tex]\rm Ksp=[Ba^{2+}][CrO_4^{2-}]\\\\1.1\;\times\;10^{-12}=0.2\;[CrO_4^2^-]\\\\CrO_4^2^-=6.0\;\times\;10^{-10}[/tex]
The concentration of chromate ions in the solution has been [tex]6.0\;\times\;10^{-10}\;\text M[/tex].
The concentration of silver ions in the solution has been:
[tex]\rm Ksp=[Ag^+]^2[CrO_4^2^-]\\\\1.1\;\times\;10^{-12}=[Ag^+]^2\;\times\;6.0\;\times\;10^{-10}\\\\Ag^+=4.2\;\times\;10^-^2\;M[/tex]
The concentration of silver ions in the solution when the barium chloride starts to precipitate is 0.0042 M.
Learn more about solubility equilibrium, here:
https://brainly.com/question/13614239
