20. A saturated solution of Agl contains 0.05 M KNO3 The Ks for Agl is 8.3 x 10-

Question

20. A saturated solution of Agl contains 0.05 M KNO3 The Ks for Agl is 8.3 x 10-17 a. As a first approximation, assume that the ionic strength of this solution is dominated by 0.05 M KNO3 Calculate . b. Using the EDH, calculate Yag+ and c. Taking activities into account, calculate [Agrl and [I] in this solution. d. Was the approximation in Q20a valid? It not, how could you amend your ionic strength calculation? 21. In this activity we considered KNO, solutions in the range 1.0x101 to 1.0x105 M. The Extended Debye-Hücke equation is a reasonable way to compute activity coefficients for 0.1 M. Suppose we have AgBr(s) ed in a 1.0 M solution of KNOg. At this ionic strength will the activity coefficients be closer to or furthe from 1 than they are for the lower concentration KNO, solutions?

Explanation / Answer

total concentration of dissolved iodine species (iodide (I-); Ag+) increases as we add “inert” KNO3to the solution. When we add inert ions like KNO3 to the solution in higher concentration , inert ion “shields” the ions in the equilibrium with counter-cations and counter-anions in their “ionic atmosphere” and decreases the tendency of the ion come together to form insoluble salt of AgI. We also can say that in presence of KNO3 attraction between Ag+ and I- decreses.

µ is a measure of the total ionic concentration in solution: but ions with more charge are counted more due to stronger electrostatic interactions with other ions

µ = 1/2 CiZi 2

where Ci is conc. of ith species

Zi is the charge on ith species

Now we know that AgI solution is saturate and at saturation Qsp =Ksp

Qsp =Ksp = [Ag+][I-]= 8.3×10-17

Therefore concentration of AgI = square root of 8.3×10-17=8.3×10-17

Concentration of [Ag+] =[I-]= 9.110 × 10-9 M

µ = 1/2 ([Ag+]ZAg2+ [I-]ZI2) = 1/2 (9.110 × 10-9 (1)2 + 9.110 × 10-9 (-1)2) = 9.110 × 10-9 M

Since we know the ionic strength of the solution we can find activity coefficient (i)of ions in that solution,

activity = ai= i[i]………. activity is simply the concentration of the species times the activity coefficient

Now, Ksp(AB) = [A][B]

Log A= 0.51×z2A×/1+3.3×A×

where zA is the ion’s charge,

A is the effective diameter of the hydrated ion in nanometers

µ is the solution’s ionic strength, and 0.51 and 3.3 are constants appropriate for an aqueous solution at 25oC

For Silver A = 0.25

For Iodine A = 0.3

Therefore now Log Ag= [0.51×z2A×]/[1+3.3×A× ]

Activity coefficient(Ag)= Log Ag= [0.51 × (1)2 × 9.110 × 10-9] / [1+3.3×0.25×9.110 × 10-9]= 0.9998

Activity coefficient (I)=Log I=[0.51 × (-1)2 × 9.110 × 10-9] / [1+3.3×0.3×9.110 × 10-9]= 0.9998

Activity of Silver Ion(Ag+) = Ag+ × [Ag+]= 0.9998 × 9.110 × 10-9 = 9.1081 × 10-9 M

Activity of Iodine Ion(I-) = I- × [I-]=0.9998 × 9.110 × 10-9 = 9.1081 × 10-9 M

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