6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...

Question

Question

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

science chemistry

Add a comment Improve this question Transcribed image text

Answer 1

Answer #1

Step 1

Moles of I^- present in 20 mL solution = MV

0.08 0,02 1.6 * 103moil

Mole of Ag^+ in 6 mL = 0.1 * 0.006 = 0.6 * 10^-^3 mol

AgI get precipitated and the remaining mol of I^- = (1.6 * 10^-^3) - (0.6 * 10^-^3) = 1.0 * 10^-^3 mol

K_s_p = [Ag^+ ] [I^-]

8.3 * 10^-^1^7 = [Ag^+ ] (10^-^3)

g8.3 10-14

Pag = - log [Ag^+ ] = - log (8.3 * 10^-^1^4) = 13.08

Thus, Pag value is 13.08

Moles of Cl^- present in 20 mL solution = MV

= 0.08 * 0.02 = 1.6 * 10^-^3 mol

Mole of Ag^+ in 6 mL = 0.1 * 0.006 = 0.6 * 10^-^3 mol

AgCI get precipitated and the remaining mol of CI^- = (1.6 * 10^-^3) - (0.6 * 10^-^3) = 1.0 * 10^-^3 mol

K_s_p = [Ag^+ ] [CI^-]

1.77 * 10^-^1^0 = [Ag^+ ] (10^-^3)

[Ag^+ ] = 1.77 * 10^-^7

Pag = - log [Ag^+ ] = - log (1.77 * 10^-^7) = 6.74

Thus, Pag value is 6.74

Step 2

Mole of Ag^+ in 16 mL = 0.1 * 0.016 = 1.6 * 10^-^3 mol

AgI get precipitated and the remaining mol of I^- = (1.6 * 10^-^3) - (1.6 * 10^-^3) = 0.0 mol

[Ag^+ ] = [I^-] K_s_p = [Ag^+ ] [I^-]

8.3 * 10^-^1^7 = [Ag^+ ]^2

[Ag^+ ] = 0.93 * 10^-^8

Pag = - log [Ag^+ ] = - log (0.93 * 10^-^8) = 8.03

Thus, Pag value is 8.03

Mole of Ag^+ in 16 mL = 0.1 * 0.016 = 1.6 * 10^-^3 mol

AgCI get precipitated and the remaining mol of I^- = (1.6 * 10^-^3) - (1.6 * 10^-^3) = 0.0 mol

[Ag^+ ] = [CI^-]

K_s_p = [Ag^+ ] [CI^-]

1.77 * 10^-^1^0 = [Ag^+ ]^2

[Ag^+ ] = 1.33 * 10^-^5

Pag = - log [Ag^+ ] = - log (1.33 * 10^-^5) = 4.86

Thus, Pag value is 4.86

Step 3

Mole of Ag^+ in 26 mL = 0.1 * 0.026 = 2.6 * 10^-^3 mole

AgI get precipitated and the extra mol of Ag^+ = (2.6 * 10^-^3) - (1.6 * 10^-^3) = 10^-^3 mol

[Ag^+ ] = M /V

= (10^-^3) /(0.02 + 0.026)

= 2.173 * 10^-^2

Pag = - log [Ag^+ ] = - log (2.173 * 10^-^2) = 1.68

Thus, Pag value is 1.68

Mole of Ag^+ in 26 mL = 0.1 * 0.026 = 2.6 * 10^-^3 mole

AgCI get precipitated and the extra mol of Ag^+ = (2.6 * 10^-^3) - (1.6 * 10^-^3) = 10^-^3 mol

[Ag^+ ] = M /V = (10^-^3) /(0.02 + 0.026) = 2.173 * 10^-^2

Pag = - log [Ag^+ ] = - log (2.173 * 10^-^2) = 1.68

Thus, Pag value is 1.68

Titration curve can be drawn using volume of AgNO_3 used for titration Vs Pag value obtained.

Add a comment

Answer 2

Similar Homework Help Questions

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...
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 =
Consider the titration of 25.00 mL of 0.0925 M KI with 0.0695 M AgNO3. Calculate pAg+...

Consider the titration of 25.00 mL of 0.0925 M KI with 0.0695 M AgNO3. Calculate pAg+ at the following volumes of added AgNO3: (a) 30.00 mL; (b) Ve; (c) 35.27 mL. Ksp (AgI) = 8.3×10-17

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+=
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 solution contains 0.020 mole each of I”, Br, and Cl. When the solution is mixed...

A solution contains 0.020 mole each of I”, Br, and Cl. When the solution is mixed with 200. mL of 0.28 M AgNO3, what mass of AgCl(s) precipitates out, and what is Ag+)? Assume no volume change AgI: Ksp = 1.5 x 10-16 AgBr : Ksp = 5.0 x 10-18 AgCl : Ksp = 1.6 x 10-10 00 Mass of AgCl(s) = [18+] - M

A 100.00mL solution of 0.100 M NaCl was titrated with 0.100 M AgNO3. A SCE indicator electrode (E+=0.241V) was the anode...

A 100.00mL solution of 0.100 M NaCl was titrated with 0.100 M AgNO3. A SCE indicator electrode (E+=0.241V) was the anode and an Ag wire was the cathode for this precipitation titration. Calculate the voltage after the addition of 65 mL of AgNO3 given that Ksp(AgCl)=1.8x10^(-10). Answer=0.081 V. Please show steps, thank you! :)
A solution is 0.10 M Cl and 0.10 M I- . Silver nitrate is slowly added...

A solution is 0.10 M Cl and 0.10 M I- . Silver nitrate is slowly added to precipitate the silver halide. Ksp(AgCl) = 1.77 x 10-10; Ksp(AgI) = 8.52 x 10-17 (A) Which solid will precipitate first? (B) What will be the [Ag+ ] when the first solid begins to precipitate? (C) What [Ag+ ] is required in order to precipitate AgCl? (D) What will be the [I- ] when the AgCl starts to precipitate? (E) What percentage of the...
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....

A chemist titrates 25.0 mL of 0.00100 M I- solution with 0.000500 M Ag+ . It...

A chemist titrates 25.0 mL of 0.00100 M I- solution with 0.000500 M Ag+ . It takes 50.0 mL of titrant to reach the equivalence point. The Ksp= 8.3 x 10-17 for AgI solid. a) What is the [I-] value when the added titrant volume is 48.0 mL? b) What is the pAg+ value when the added titrant volume is 48.0 mL?

6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...

Homework Answers

Add Answer to:
6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...

Post as a guest

Earn Coins

Perform the calculations needed to generate a titration curve for 50.00 ml of a 0.0500 M...

A 25.00 mL solution of 0.08610 M Nal is titrated with 0.05000 M AgNO . Calculate...

Consider the titration of 25.00 mL of 0.0925 M KI with 0.0695 M AgNO3. Calculate 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...

Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of compound...

A solution contains 0.020 mole each of I”, Br, and Cl. When the solution is mixed...

A 100.00mL solution of 0.100 M NaCl was titrated with 0.100 M AgNO3. A SCE indicator electrode (E+=0.241V) was the anode...

A solution is 0.10 M Cl and 0.10 M I- . Silver nitrate is slowly added...

At 25 ∘C, a titration of 15.00mL of a 0.0360 M AgNO3 solution with a 0.0180...

A chemist titrates 25.0 mL of 0.00100 M I- solution with 0.000500 M Ag+ . It...

6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...

Homework Answers

Add Answer to: 6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...

Post as a guest

Earn Coins

Add Answer to:
6A. Construct the titration curve for a 20.00 m, mixture solution of 0.0800 M in I...