Common Ion Effect | MCAT Organic Chemistry Prep
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Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach the common ion effect of solutions. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!
When an electrolyte dissolves in solution, its associated ion pair dissociates into a separate cation and anion. The equilibrium concentration of ions dissolved in solution is denoted by the solubility product constant (Ksp). However, what would happen if extra ions were added to a solution at equilibrium? This effect, the common ion effect, will be demonstrated through two examples.
Example 1
A solution of dissolved CaCO3 is at equilibrium. What would happen to the amount of dissolved CaCO3 if CaCl2 is added to the solution?
Applying Le Chatelier's principle, a secondary equilibrium will begin, where CaCl2 will dissolve and dissociate to form calcium ions and chloride ions that will freely interact with solution. Both equilibriums present in the solution share a common ion, calcium. The dissolution of CaCl2 increases the total amount of calcium ion present in solution. Remembering that the dissolution of CaCO3 is at equilibrium, an increase in calcium will increase the ion product (Q) of CaCO3 past its Ksp. When Q greater than Ksp, excess solute precipitates out of solution. In this case, excess Ca2+ and will precipitate, forming CaCO3 until Q = Ksp. The take home point here is that addition of a common ion will decrease solubility, potentially forcing some excess ion to form a precipitate.
Example 2
A solution of dissolved CaCO3 is at equilibrium. What would happen to the amount of dissolved CaCO3 if HCl is added to the solution?
Recognizing that HCl is a strong acid and that CO3 is a weak base, we can predict that the two are going to undergo a neutralization reaction. As the hydrogen chloride protonates the bicarbonate, carbonate will be removed from solution. If ions are removed from equilibrium, we will reach a situation where Q less than Ksp. When Q less than Ksp, the reaction is unsaturated, and fewer ions are dissolved in solution than will be at equilibrium. To increase the ion product (Q), calcium carbonate will dissociate further into calcium and bicarbonate ions, until Q = Ksp. The take home point from this is example is that removal of a common ion will increase solubility.
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When an electrolyte dissolves in solution, its associated ion pair dissociates into a separate cation and anion. The equilibrium concentration of ions dissolved in solution is denoted by the solubility product constant (Ksp). However, what would happen if extra ions were added to a solution at equilibrium? This effect, the common ion effect, will be demonstrated through two examples.
Example 1
A solution of dissolved CaCO3 is at equilibrium. What would happen to the amount of dissolved CaCO3 if CaCl2 is added to the solution?
Applying Le Chatelier's principle, a secondary equilibrium will begin, where CaCl2 will dissolve and dissociate to form calcium ions and chloride ions that will freely interact with solution. Both equilibriums present in the solution share a common ion, calcium. The dissolution of CaCl2 increases the total amount of calcium ion present in solution. Remembering that the dissolution of CaCO3 is at equilibrium, an increase in calcium will increase the ion product (Q) of CaCO3 past its Ksp. When Q greater than Ksp, excess solute precipitates out of solution. In this case, excess Ca2+ and will precipitate, forming CaCO3 until Q = Ksp. The take home point here is that addition of a common ion will decrease solubility, potentially forcing some excess ion to form a precipitate.
Example 2
A solution of dissolved CaCO3 is at equilibrium. What would happen to the amount of dissolved CaCO3 if HCl is added to the solution?
Recognizing that HCl is a strong acid and that CO3 is a weak base, we can predict that the two are going to undergo a neutralization reaction. As the hydrogen chloride protonates the bicarbonate, carbonate will be removed from solution. If ions are removed from equilibrium, we will reach a situation where Q less than Ksp. When Q less than Ksp, the reaction is unsaturated, and fewer ions are dissolved in solution than will be at equilibrium. To increase the ion product (Q), calcium carbonate will dissociate further into calcium and bicarbonate ions, until Q = Ksp. The take home point from this is example is that removal of a common ion will increase solubility.
MEDSCHOOLCOACH
To watch more MCAT video tutorials like this and have access to study scheduling, progress tracking, flashcard and question bank, download MCAT Prep by MedSchoolCoach
IOS Link: https://play.google.com/store/apps/de...
Apple Link: https://apps.apple.com/us/app/mcat-pr...
#medschoolcoach #MCATprep #MCATstudytools
4 سال پیش
در تاریخ 1399/08/06 منتشر شده
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