Aluminumcan form two complex ions with fluoride, AlF63–(Kf= 2.4x10^4), which forms in solutions with [F–] > 1.5M and AlF4–(Kf= 2.0x10^8), which forms when [F–] is between 0.5 M and 1.5M. 100 mL of 0.20 M Al(NO3)3is combined with 500 mL of a pH 4.25 HF/NaF buffer (total concentration of 2.0M). a.Calculate [Al3+], [F–], and the [complex ion] at equilibrium. b.Given that the sodium salt of the complex ion has a solubility of 0.042 g/100 mL, will the sodium salt precipitate from the equilibrium mixture? Why or why not? If AlF63–forms, the sodium salt would be Na3AlF6; if AlF4–forms, the sodium salt would be NaAlF4.
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
Aluminumcan form two complex ions with fluoride, AlF63–(Kf= 2.4x10^4), which forms in solutions with [F–] > 1.5M and AlF4–(Kf= 2.0x10^8), which forms when [F–] is between 0.5 M and 1.5M. 100 mL of 0.20 M Al(NO3)3is combined with 500 mL of a pH 4.25 HF/NaF buffer (total concentration of 2.0M).
a.Calculate [Al3+], [F–], and the [complex ion] at equilibrium.
b.Given that the sodium salt of the complex ion has a solubility of 0.042 g/100 mL, will the sodium salt precipitate from the equilibrium mixture? Why or why not? If AlF63–forms, the sodium salt would be Na3AlF6; if AlF4–forms, the sodium salt would be NaAlF4.
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