A mixture of metal ions in a solution can be separated by precipitation with anions such as Cl−Cl−\ce{Cl-}, Br−Br−\ce{Br-}, SO2−4SO42−\ce{SO4^2-}, CO2−3CO32−\ce{CO3^2-}, S2−S2−\ce{S^2-}, Cr2O2−4Cr2O42−\ce{Cr2O4^2-}, PO2−4PO42−\ce{PO4^2-}, OH−OH−\ce{OH-} etc.
What is the fundamental principle that allows these anions to separate different metal ions?"
Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure 4.8.14.8.1\PageIndex{1}. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
"The text explains that adding ammonia to a copper solution changes the color from light blue to deep blue-violet. Why does this color change occur, and what does it tell us about the chemical species formed?"
Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution.
Why does the pH of a salt solution depend on the strength of the acid and base that formed the salt? For example, why does a salt like potassium cyanide produce a basic solution, while ammonium chloride produces an acidic solution, even though both salts are dissolved in water?
Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
Why does the percent Ionization of a weak acid decrease as the Initial concentration increases?
lowering the concentration of one of the reactants would cause QQQ to be larger than K)
How does Q determine the directional shift when the reactant is added? Does K affect it too if changed?
(2.1.1)N2O4(g)colorless⇌krkf2NO2(g)red−brown\underset{colorless }{\ce{N2O4 (g)}} \ce{ <=>[k_f][k_r] } \underset{red-brown }{\ce{2NO2(g)}}\label{Eq1} The double arrow indicates that both the forward reaction N2O4(g)−→kf2NO2(g)(2.1.2)(2.1.2)N2O4(g)→kf2NO2(g)\ce{N2O4 (g) ->[k_f] 2NO2(g)} \label{eq1B} and reverse reaction 2NO2(g)−→krN2O4(g)(2.1.3)(2.1.3)2NO2(g)→krN2O4(g)\ce{2NO2(g) ->[k_r] N2O4 (g) } \label{eq1C}
How is the reaction between NO2 and N2O4 considered reversible? What does it demonstrate?