Аналитическая химия - Помогите решить задачи по теме растворимость
Помогите! Ниже приведен 3 вопроса по аналитической химии сначала на английском (оригинал), а потом перевод на русский
1. Fe (III) can be precipitated from acidic solution by addition of OH- to form Fe(OH)3. At what concentration of OH- will the concentration of Fe (III) be reduced to 1.0*10-10 M? If Fe (II) is used instead, what concentration of OH- is necessary to reduce the Fe (II) concentration to 1.0*10-10 M?
Ksp (Fe(OH)3)=1.6*10^-39; Ksp (Fe(OH)2)=7.9*10^-16;
2. Common ion effect. Express the solubility of AgIO3 in 10.0 mM KIO3 as a fraction of its solubility in pure water.
3. Separation by precipitation. What concentration of carbonate must be added to 0,10 M Zn2+ to precipitate 99,90% of the Zn2+ ?
1. Fe (III) может быть осажден из кислотного раствора добавлением OH- с образованием Fe (OH)3. При какой концентрации OH- концентрация Fe (III) будет снижена до 1,0 * 10-10 М? Если вместо него используется Fe (II), то какая концентрация OH- необходима для снижения концентрации Fe (II) до 1,0*10-10 М?
2. Эффект одноименного (общего) иона. Выразить растворимость AgIO3 в 10,0 мМ KIO3 как долю его растворимости в чистой воде.
3. Разделение осаждением. Какую концентрацию карбоната необходимо добавить к 0,10 М Zn2+ для осаждения 99,90% Zn2+ ?
1. To find the concentration of OH- required to reduce the concentration of Fe (III) to 1.0*10^-10 M, we need to consider the solubility product constant (Ksp) of Fe(OH)3.
The equation for the reaction between Fe (III) and OH- is:
Fe (III) + 3OH- -> Fe(OH)3
The expression for the solubility product of Fe(OH)3 is given as:
Ksp (Fe(OH)3) = [Fe(III)][OH-]^3 = 1.6*10^-39
To find the concentration of OH- necessary, we can rearrange the equation:
[OH-] = (Ksp (Fe(OH)3) / [Fe(III)])^(1/3)
Substituting the values we have:
[OH-] = (1.6*10^-39 / 1.0*10^-10)^(1/3) = 1.6*10^-10 M
Therefore, the concentration of OH- required to reduce the concentration of Fe (III) to 1.0*10^-10 M is 1.6*10^-10 M.
If Fe (II) is used instead, we need to consider the solubility product constant of Fe(OH)2.
The equation for the reaction between Fe (II) and OH- is:
Fe (II) + 2OH- -> Fe(OH)2
The expression for the solubility product of Fe(OH)2 is given as:
Ksp (Fe(OH)2) = [Fe(II)][OH-]^2 = 7.9*10^-16
Using the same approach as above, we can calculate the concentration of OH- needed:
[OH-] = (Ksp (Fe(OH)2) / [Fe(II)])^(1/2)
[OH-] = (7.9*10^-16 / 1.0*10^-10)^(1/2) = 2.8*10^-3 M
Therefore, the concentration of OH- required to reduce the concentration of Fe (II) to 1.0*10^-10 M is 2.8*10^-3 M.
2. In the scenario of the common ion effect, we need to express the solubility of AgIO3 in 10.0 mM KIO3 as a fraction of its solubility in pure water.
The common ion effect states that adding a common ion to a solution reduces the solubility of a slightly soluble compound. In this case, the common ion is IO3-.
Let's assume the solubility of AgIO3 in pure water is represented by [AgIO3]pure.
The solubility of AgIO3 in the presence of 10.0 mM KIO3 is represented by [AgIO3]10.0 mM.
According to the common ion effect, we have the relationship:
[AgIO3]10.0 mM = [AgIO3]pure * [IO3-] (where [IO3-] is the concentration of IO3- in 10.0 mM KIO3)
Therefore, the solubility of AgIO3 in 10.0 mM KIO3 as a fraction of its solubility in pure water is:
([AgIO3]10.0 mM / [AgIO3]pure) = [IO3-]
3. In order to precipitate 99.90% of the Zn2+ from a 0.10 M Zn2+ solution by the addition of carbonate, we need to consider the solubility product constant (Ksp) of the precipitate formed, which is ZnCO3.
The equation for the reaction between Zn2+ and CO32- is:
Zn2+ + CO32- -> ZnCO3
To find the concentration of carbonate necessary, we can rearrange the equation for the solubility product of ZnCO3:
Ksp (ZnCO3) = [Zn2+][CO32-] = (0.10 - 0.10*0.999) * (0.10 - 0.10*0.999) = 1.0*10^-10
Assuming the reaction goes to completion, we can simplify the equation:
(0.10*0.001)^2 = 1.0*10^-10
0.0001 = 1.0*10^-10
1 = 1.0*10^-6
Therefore, the concentration of carbonate necessary to precipitate 99.90% of the Zn2+ is 1.0*10^-6 M.
I hope this helps!