± The Common-Ion Effect The common-ion effect and solubility The common-ion effe

Question

± The Common-Ion Effect The common-ion effect and solubility The common-ion effect is an application of Le Châtlier's principle, which states that an equilibrium system that is stressed will work to alleviate that stress and re-establish equilibrium The solubility of a slightly soluble salt can be greatly affected by the addition of a soluble salt with a common ion, that is, with one of the ions in the added soluble salt being identical to one of the ions of the slightly soluble salt. The general result of the addition of the common ion is to greatly reduce the solubility of the slightly soluble salt. In other words, the addition of the common ion results in a shift in the equilibrium of the slightly soluble salt Part A Mg(OH) 2 is a sparingly soluble salt with a solubility product constant, Ksp , of 5.61 × 10-11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. Calculate the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.180 mol L NaOH solution. Express your answer numerically to three significant figures 2.31 10 Submit Hints My Answers Give Up Review Part Incorrect; One attempt remaining; Try Again The common-ion effect and buffer systems A buffer is a mixed solution of a weak acid or base, combined with its conjugate. Note that this can be understood essentially as a common-ion problem: The conjugate is a common ion added to an equilibrium system of a weak acid or base. The addition of the conjugate shifts the equilibrium of the system to relieve the stress of the added concentration of the common ion. In a solution consisting of a weak acid or base, the equilibrium shift also results in a pH shift of the system It is the presence of the common ion in the system that results in buffering behaviour, because both added H+ or OH ions can be neutralized

Explanation / Answer

Mg(OH)2 (s) <--> Mg2+ (aq) + 2OH- (aq)

Ksp = [Mg2+] [OH-]^2

5.61 x 10^ -11 = (S) ( 2S) ^2    ( S is solubility)

S = 0.00024

Now in presence of 0.18NaoH we have OH- contribution from NaoH = 0.18 M

Ksp = 5.61 x 10^ -11 = (S) ( 2S + 0.18) ^2

5.61 x 10^ -11 = S ( 0.18^2)      ( since Ksp is very low we get low value of S , hence 2S value negligible compared to 2S    and hence 2S + 0.18 = 2S)

S = 1.73 x 10^ -9

Now ratio of solubilities = 0.00024 / ( 1.73x10^-9) = 138728

2) pa = 4.77 , Ka = 10^ -4.77 = 1.7 x 10^ -5

Initially we had 0.2 M citric acid   let acid be HA

HA (aq) <--> H+ (aq) + A-(aq)

at equilibrium HA] = 0.2-X , [H+] =[A-] = X

Ka = [H+] [A-] / [HA]

1.7 x 10^ -5 = X^2 / ( 0.2-X)

X = [H+] = 0.00183543 ,

pH = -log [H+] = -log ( 0.00183543) = 2.736

now 0.13 M citrate added

we use Buffer equation

pH = pka + log [citrate] / [citric acid]

pH = 4.77 + log ( 0.13 / 0.2)

pH = 4.583

pH change = 4.583 - 2.736 = 1.847

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