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
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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
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![**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}^+ = -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F97e76988-7abc-4229-a25b-aa67402e03fe%2F3cabea61-85b9-47e0-8259-71296e885d92%2Fqq925kp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**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}^+ = -
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