![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.
Add Answer to:
6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...
Perform the calculations needed to generate a titration curve for 50.00 ml of a 0.0500 M...
Perform the calculations needed to generate a titration curve for 50.00 ml of a 0.0500 M NaCl solution titrated with 0.1000 M AgNO3 . Note for AgCl the Ksp = 1.82×10-10 . (i) Calculate pAg when 10.00 ml of AgNO3 is added. (ii) Calculate pAg when 25.00 ml of AgNO3 is added. (iii) Calculate pAg when 26.00 ml of AgNO3 is added. Given the solubility products above, show (by calculation) which of the two compounds concerned has the greater solubility...
Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of compound...
Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of compound A with a 0.2000 M solution of compound B in the following table. For each titration, calculate the pH after the addition of 0.00, 12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45.00, 49.00, 50.00, 51.00, and 60.00 mL of com- pound B. A (a) H2SO3 (b) ethylenediamine (c) H2SO4 B NaOH HCl NaOH
A 25.00 mL solution of 0.08610 M Nal is titrated with 0.05000 M AgNO . Calculate...
A 25.00 mL solution of 0.08610 M Nal is titrated with 0.05000 M AgNO . Calculate pAg+ following the addition of the given volumes of AgNO,. The Ksp of Agl is 8.3 x 10-'7. 37.40 mL pAgt = V pAgt = 47.90 mL pAgt =
A 25.00 mL solution of 0.08680 M NaI is titrated with 0.05120 M AgNO3 . Calculate...
A 25.00 mL solution of 0.08680 M NaI is titrated with 0.05120 M AgNO3 . Calculate pAg+ following the addition of the given volumes of AgNO3 . The ?sp of AgI is 8.3×10^−17 36.00 mLpAg+ = ? ?e pAg+ = ? 47.40 mL pAg+ = ?
A 25.00 mL solution of 0.08050 M NaI is titrated with 0.05000 M AgNO3 . Calculate pAg+ following the addition of the giv...
A 25.00 mL solution of 0.08050 M NaI is titrated with 0.05000 M AgNO3 . Calculate pAg+ following the addition of the given volumes of AgNO3 . The ?sp of AgI is 8.3×10−17 . 35.80 mLpAg+= ?epAg+= 47.90 mLpAg+=
A 20.00-mL aqueous solution containing tin(II) ions with a concentration of 0.0850 M is titrated with...
A 20.00-mL aqueous solution containing tin(II) ions with a concentration of 0.0850 M is titrated with a standard aqueous solution of cerium(IV) ions having a concentration of 0.1100 M. Write a balanced chemical equation for the titration reaction and then find the volume, in mL, of the titrant that is required to reach the stoichiometric point of the titration. b. If the titration were followed potentiometrically using a platinum electrode and a saturated Ag/AgCl electrode, find the following voltages in...
At 25 ∘C, a titration of 15.00mL of a 0.0360 M AgNO3 solution with a 0.0180...
At 25 ∘C, a titration of 15.00mL of a 0.0360 M AgNO3 solution with a 0.0180 M NaI solution is conducted within the cell SCE || titration solution | Ag(s) For the cell as written, what is the potential after the addition of each volume of NaI solution? The reduction potential for the saturated calomel electrode is ?=0.241 V. The standard reduction potential for the reaction Ag+ + e− ⟶Ag(s) is ?∘=0.79993 V. The solubility constant of AgI is ?sp=8.3×10−17....
Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant. (a) 15.00 mL: pH = (b) 20.4...
Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4), with 0.1000 M HCl solution after the following additions of titrant. (a) 15.00 mL: pH = (b) 20.40 mL: pH = (c) 29.00 mL: pH =
QUESTION 1 Which of the following solutions is a buffer? A mixture of HF and HBr A...
QUESTION 1 Which of the following solutions is a buffer? A mixture of HF and HBr A mixture of H2SO4 and Na2SO4 A mixture of NaI and HI A mixture of NaF and HF 1 points QUESTION 2 What is the [A-]/[HA] ratio in a buffer made of HA and its conjugate base, A-, and has a pH of 8.00 ? The pKa of HA is 6.00 1:1 1:10 1:100 10:1 100:1 1 points QUESTION 3 If the molar...
A 25.00 mL solution of 0.08610 M Nal is titrated with 0.05000 M AgNO . Calculate pAg+ following the addition of the given volumes of AgNO,. The Ksp of Agl is 8.3 x 10-'7. 37.40 mL pAgt = V pAgt = 47.90 mL pAgt =
Most questions answered within 3 hours.
-
Where is the error in this code sequence?
String s1 = "Hello";
String s2 = "ello";...
asked 10 months ago -
Financial data for Joel de Paris, Inc., for last year
follow:
Joel de Paris, Inc.
Balance...
asked 10 months ago -
Consider this reaction:
Al2(SO4)3 (aq)+ BaCl3
(aq) Al2Cl6 (aq)- +
3BaSO4(s) . What is the...
asked 10 months ago -
Suppose that Savneet is considering increasing her
recent random sample from 20 car rentals to 40...
asked 10 months ago -
Trucks arrive at an unloading terminal at an average rate of 120
per hour.
Trucks arrive...
asked 10 months ago -
Why are methanol and ethanol completely soluble in water while
octanol is not very little soluble....
asked 10 months ago -
A facilities manager at a university reads in a research report
that the mean amount of...
asked 10 months ago -
When the CuSO4 is rehydrated by adding water to the anhydrous
compound, is this an endothermic...
asked 10 months ago -
A ray of sunlight is passing from diamond into crown glass; the
angle of incidence is...
asked 10 months ago -
A block of mass 0.249 kg is placed on top of a light, vertical
spring of...
asked 10 months ago -
how do the kidneys compensate in the presences of acidosis
a) trigger hyperventilate
b) reserve acid...
asked 10 months ago -
Question 501 pts
The rental rate of capital to the firm increases. Which of the
following...
asked 10 months ago