We need to increase the concentration of common ion first, in order to promote the common ion effect
<h3>What is the Common ion effect?</h3>
It is an effect that suppresses the dissociation of salt due to the addition of another salt having common ions.
For example, a saturated solution of silver chloride in equilibrium has Ag⁺ and Cl⁻ . Sodium Chloride is added to the solution and has a common ion Cl⁻. As a result, the equilibrium shifts to the left to form more silver chloride. Thus, solubility of AgCl decreases.
The Equilibrium law states that if a process is in equilibrium and is subjected to a change
- in temperature,
- pressure,
- the concentration of reactant or product,
then the equilibrium shifts in a particular direction, according to the condition.
Thus, an increase in the concentration of common ion promotes the common ion effect.
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We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal
2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008
When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4%
For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal
When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂
IMixing combined with skip voting
