**Exercise**: Precipitation titration is a volumetric method based upon the formation of slightly soluble precipitate:\[ \text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \] \[ k_{sp} = 1.6 \times 10^{-10} \]Construct a titration curve by plotting \(\text{pAg}^+\) vs. \(V_{\text{AgNO}_3}\) added.| \(V_{\text{AgNO}_3}\) (ml) | \([\text{Ag}^+]\) | \(\text{pAg}^+\) ||----------------------------|-----------------|----------------|| 0.0 | | || 5 | | || 10.0 | | || 15 | | || 20 | | || 25 | | || 30 | | || 35 | | || 40 | | || 45 | | || 50 | | |**Hints**:(a) Before equivalence point: \([ \text{Cl}^-] = [\text{NaCl}] = \frac{[\text{NaCl}] \times V_{\text{NaCl}} - [\text{AgNO}_3] \times V_{\text{AgNO}_3}}{V_{\text{NaCl}} + V_{\text{AgNO}_3}}\)\[ [\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} \](b) At equivalence point (moles of AgNO₃ = moles of NaCl): \([\text{Ag}^+] = \sqrt{k_{sp}}\)(c) After the equivalence point:\[ [\text{Ag}^+] = \frac{[\text{AgNO}_3] \times V_{\text{AgNO}_3}}{50 + V_{\text{AgNO}_3}} \](d) \(\text{pAg}^+ = -

Using the concept of precipitation titration, pAg can be calculated at the equivalence point of the…

Exercise Precipitation titration is a volumetric method based upon the formation of slightly soluble Precipitate: AgNO3(aq) + NaCl(aq) AgCl(s) Construct a titration curve by plotting pAg vs. VAgNO3 added VAgNO3 (ml) [Ag¹] pAg* 0.0 5 10.0 15 Hints: 20 25 30 a 35 40 45 50 oo (a) Before equivalence point: d. b. 1.43 1.30 K sp [CI-] (b) At equivalence point (moles of AgNO3 = moles of NaCl): [Ag*]=√ksp (c) After the equivalence point: Ag (aq) 4.90 [CI]=[NaCl] = 7.00 e. 7.65 [Ag*] = (d) pAg = -log[Ag*] (e) Set the values of Ksp, [NaC1], VNaCI and [AgNO3] as reference cell. What is pAg at the equivalence point of titration? + Cr [AgNO3]*VA AgNO, 50+ V AgNO, (aq) [Ag] = Ksp [NaCl]*V Naci -[AgNO3]*VA AgNO V Naci+V AgNO = 1.6 10-10 titrant: AgNO3, 0.IM analyte: NaC1, 0.05M, 50 ml

Exercise Precipitation titration is a volumetric method based upon the formation of slightly soluble Precipitate: AgNO3(aq) + NaCl(aq) AgCl(s) Construct a titration curve by plotting pAg vs. VAgNO3 added VAgNO3 (ml) [Ag¹] pAg* 0.0 5 10.0 15 Hints: 20 25 30 a 35 40 45 50 oo (a) Before equivalence point: d. b. 1.43 1.30 K sp [CI-] (b) At equivalence point (moles of AgNO3 = moles of NaCl): [Ag]=√ksp (c) After the equivalence point: Ag (aq) 4.90 [CI]=[NaCl] = 7.00 e. 7.65 [Ag] = (d) pAg = -log[Ag] (e) Set the values of Ksp, [NaC1], VNaCI and [AgNO3] as reference cell. What is pAg at the equivalence point of titration? + Cr [AgNO3]VA AgNO, 50+ V AgNO, (aq) [Ag] = Ksp [NaCl]V Naci -[AgNO3]VA AgNO V Naci+V AgNO = 1.6 10-10 titrant: AgNO3, 0.IM analyte: NaC1, 0.05M, 50 ml

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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