![6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I and 0.0800M in Cl titrated with 0.1000 M AgNO3. Calculate the pAg values of the titration solution atter addition of 6.00mL, 16.00 mL, and 26.00 mL of 0.1000 M AgNO solution. 18) Kp AgCl = 1.77 × 10-10ー[Ag+][Cl] Ksp Agl 8.3 x 101-AgI](http://img.homeworklib.com/questions/225836c0-a17b-11ea-9c5d-35327ebbcaaa.png?x-oss-process=image/resize,w_560)
Homework Answers
Step 1
Moles of I^- present in 20 mL solution = MV
Mole of Ag^+ in 6 mL 
AgI get precipitated and the remaining mol of 
![K_s_p = [Ag^+ ] [I^-]](http://img.homeworklib.com/questions/3a0ae270-af30-11ea-af67-7dd0ceae66c9.png?x-oss-process=image/resize,w_560)
![8.3 * 10^-^1^7 = [Ag^+ ] (10^-^3)](http://img.homeworklib.com/questions/3a71c680-af30-11ea-ba85-37c7c61126d8.png?x-oss-process=image/resize,w_560)
![Pag = - log [Ag^+ ] = - log (8.3 * 10^-^1^4) = 13.08](http://img.homeworklib.com/questions/3b36b7f0-af30-11ea-9558-23c6658c0edd.png?x-oss-process=image/resize,w_560)
Thus, Pag value is 13.08
Moles of Cl^- present in 20 mL solution = MV
Mole of Ag^+ in 6 mL
AgCI get precipitated and the remaining mol of CI^- 
![K_s_p = [Ag^+ ] [CI^-]](http://img.homeworklib.com/questions/3cacec40-af30-11ea-8eb7-3f84940abe17.png?x-oss-process=image/resize,w_560)
![1.77 * 10^-^1^0 = [Ag^+ ] (10^-^3)](http://img.homeworklib.com/questions/3d0bfdf0-af30-11ea-bec1-13cfc8d1ffc6.png?x-oss-process=image/resize,w_560)
![[Ag^+ ] = 1.77 * 10^-^7](http://img.homeworklib.com/questions/3d6cf800-af30-11ea-a22a-1d26ebc2f25d.png?x-oss-process=image/resize,w_560)
![Pag = - log [Ag^+ ] = - log (1.77 * 10^-^7) = 6.74](http://img.homeworklib.com/questions/3dcd1240-af30-11ea-9dc2-cd4d5c20e4c8.png?x-oss-process=image/resize,w_560)
Thus, Pag value is 6.74
Step 2
Mole of Ag^+ in 16 mL =
AgI get precipitated and the remaining mol of 
![[Ag^+ ] = [I^-] K_s_p = [Ag^+ ] [I^-]](http://img.homeworklib.com/questions/3efca570-af30-11ea-afff-39d3a011d2d1.png?x-oss-process=image/resize,w_560)
![8.3 * 10^-^1^7 = [Ag^+ ]^2](http://img.homeworklib.com/questions/3f5f3230-af30-11ea-93ce-93db51ca8e27.png?x-oss-process=image/resize,w_560)
![[Ag^+ ] = 0.93 * 10^-^8](http://img.homeworklib.com/questions/3fc0c0a0-af30-11ea-beb5-2bfb44d909f4.png?x-oss-process=image/resize,w_560)
![Pag = - log [Ag^+ ] = - log (0.93 * 10^-^8) = 8.03](http://img.homeworklib.com/questions/4029c420-af30-11ea-a78e-aff5fb416bf7.png?x-oss-process=image/resize,w_560)
Thus, Pag value is 8.03
Mole of Ag^+ in 16 mL 
AgCI get precipitated and the remaining mol of
![[Ag^+ ] = [CI^-]](http://img.homeworklib.com/questions/415fd280-af30-11ea-9092-69ff9dcc4393.png?x-oss-process=image/resize,w_560)
![1.77 * 10^-^1^0 = [Ag^+ ]^2](http://img.homeworklib.com/questions/4221e410-af30-11ea-b4b1-fdd0c4bbcb11.png?x-oss-process=image/resize,w_560)
![[Ag^+ ] = 1.33 * 10^-^5](http://img.homeworklib.com/questions/4286a550-af30-11ea-97f9-817ad45268f9.png?x-oss-process=image/resize,w_560)
![Pag = - log [Ag^+ ] = - log (1.33 * 10^-^5) = 4.86](http://img.homeworklib.com/questions/42e6f8c0-af30-11ea-990d-f7491db64e52.png?x-oss-process=image/resize,w_560)
Thus, Pag value is 4.86
Step 3
Mole of Ag^+ in 26 mL 
AgI get precipitated and the extra mol of Ag^+ 
![[Ag^+ ] = M /V](http://img.homeworklib.com/questions/4420f870-af30-11ea-8d31-bd2dfc4dbcae.png?x-oss-process=image/resize,w_560)
![Pag = - log [Ag^+ ] = - log (2.173 * 10^-^2) = 1.68](http://img.homeworklib.com/questions/453e8360-af30-11ea-831e-e35ed0574ce3.png?x-oss-process=image/resize,w_560)
Thus, Pag value is 1.68
Mole of Ag^+ in 26 mL
AgCI get precipitated and the extra mol of Ag^+
![[Ag^+ ] = M /V = (10^-^3) /(0.02 + 0.026) = 2.173 * 10^-^2](http://img.homeworklib.com/questions/4655c900-af30-11ea-b299-a5d2a8bf6e82.png?x-oss-process=image/resize,w_560)
Thus, Pag value is 1.68
Titration curve can be drawn using volume of AgNO_3 used for titration Vs Pag value obtained.
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