the concentration of the ions present in the solution is
directly proportional to the solubility, the solubility of
the ion decreases. Further investigation can be made
by using a non-saturated solution of KCl to see whether
KCl would experience the common ion effect or not.
It is expected to see that it would not experience the
common ion effect since the salt would be completely
dissolved so there would be no equilibrium maintained.
According to the Le Châtelier's Principle Prediction of
Response to Stress which states if a system is reached
equilibrium (when the rate of forward and backward
reactions are same), the system resists the change
applied to the system to restore equilibrium. (Martin,
2015). It can be said that as NaCl is added to the system
the concentration of Cl- ions in the solution increases
and the system gives reaction by proceeding in the
backward reaction to increase the concentration of
the KCl( s ) salt. Hence the concentration of the product
decreases which result in the decrease in the solubility,
as the concentration of the reactant increases. More
addition of the common ion results in more shifting of
the equilibrium towards the product side so that less
concentration of ions present in the solution. Since References:
Lumen. (n.d.). Effect of a Common Ion on Solubility. Retrieved
from Introduction to Chemistry: https://courses.lumenlear-
ning.com/introchem/chapter/effect-of-a-common-ion-on-so-
lubility/
Martin, C. (2015). General Chemistry. New York: Cambridge
University Press.
Owen, S. (2014). Chemistry for IB Diploma. London: Cambrid-
ge University Press.
Petrucci, Harwood, Herring, & Madura. (2006). General Che-
mistry Principles and Modern Applications. Pearson Internati-
onal Edition.
Nihat Nebil BÜYÜKGÖLCİGEZLİ
12- B
IB DP Chemistry Puzzle-1 (All the words begin with A, B and C letters.)
THE CLAPPER 2018 - 2019
53
to expecting addition of KCl salt as the common added
ion to the NaCl solution to alter solubility of NaCl. Since
the KCl amount would not change the best fit line would
be straight line parallel to the x-axis hence the gradient
would be zero and the solubility would be constant.