Meeting 7 Principles: sedimentation of Ag(I) reaction Diffuculties of titration: HHard to find the suitable indicators IIn some cases (mainly in.

Presentasi berjudul: "Meeting 7 Principles: sedimentation of Ag(I) reaction Diffuculties of titration: HHard to find the suitable indicators IIn some cases (mainly in."— Transcript presentasi:

1

2 Meeting 7

3 Principles: sedimentation of Ag(I) reaction Diffuculties of titration: HHard to find the suitable indicators IIn some cases (mainly in dilute solution ), reaction rate usually slow TThe composition of sediment sometimes can not be determined Analite + Titrant  precipitation  AgNO 3  ARGENTOMETRy

4 Benefit : Determine Cl -, F -, Br - Reaction: NaCl (aq) + AgNO 3 (aq)  AgCl (s) + NaNO 3 (aq) white

5 Solubility constant (KSp) AgCl (s) Ag + (aq) + Cl - (aq) Ksp = [Ag + ] [Cl - ] ----- equillibrium Ksp > [Ag + ] [Cl - ] ----- become sediment

6 Indicators in Argentometric titration a. Formation of colorfull sediment: (Mohr methods: pH 6-10) b. Formation of colorfull complexs (Volhard methods, acid pH) c. Using adsoprtion indicators (Fajans Methods

7 Mohr Methods: Indicator  chromates Ag + + Cl - ==== AgCl (white crystal) Ag + + CrO 4 = === AgCrO 4 (red crystal surrounding AgCl surface)

8 Volhard methods: Indicator: Fe (III) Ag + + Cl - ==== AgCl (white crystal) Ag + + SCN - AgSCN(s) Fe 3+ + SCN - FeSCN 2+ (aq) End Point : white crystal in reddish solution

9 Fajans Methods: Indicator fluororescense (Fl) (AgCl).Ag -+ X - secondair layer Primary layer Cl- exceed Ag + + Cl - AgCl(s) (AgCl).Ag+Fl - HFlH + + Fl - E.P : (Pink precipitate)

10 pCl - = -log [Cl - ] = -log 10 -1 = 1,00 [Ag + ] = 0 a. Pada keadaan awal : Penyelesaian : Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 -  AgCl(s) + Na + (aq) + NO 3 - (aq)

11 b. Penambahan 10,0 mL AgNO 3 jumlah mmol NaCl setelah penambahan AgNO 3 [Cl - ] = ———————————————————————— volume campuran jumlah mmol NaCl awal – jumlah AgNO 3 yg ditambahkan [Cl - ] = ————————————————————————————— volume campuran (50,0 x 0,10) – (10,0 x 0,1) [Cl - ] = ——————————————— = 0,067 M 50,0 + 10,0 pCl = 1,17

12 c. Penambahan 49,9 mL AgNO 3 (50,0 x 0,10) – (49,9 x 0,1) [Cl - ] = ——————————————— = 1,0 x 10 -4 M 50,0 + 49,9 pCl = 1,17

13 d. Pada penambahan 50,00 mL AgCl(s)Ag + (aq) + Cl - (aq) [Ag + ][Cl - ] = Ksp, pAg + pCl = 10 [Ag + ] = [Cl - ] [Cl - ] 2 = 1x 10 -10 [Cl - ] = 1 x 10 -5 pCl - = 5,00

14 e. Penambahan 60,0 mL (60,0 x 0,10) – (50 x 0,1) [ Ag + ] = ——————————————— = 9,1 x 10 -3 M 60,0 + 50,0 pAg = 2,04

15 KURVA TITRASI

16

17 Metode Mohr : Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 -  AgCl(s) + Na + (aq) + NO 3 - (aq) Ag + (aq) + Cl - (aq) AgCl(s) Ksp = 1 x 10 -10 2Ag + (aq) + CrO 4 2- (aq)Ag 2 CrO 4 (s) Ksp = 2 x10 -12 Berapa konsentrasi K 2 CrO 4 yang diperlukan pada titik eakhir titrasi?

18 Konsentrasi K 2 CrO 4 yang diperlukan pada titik akhir titrasi: Kelarutan Ag 2 CrO 4 = 8,4 x 10 -5 mol/L Kelarutan AgCl = 1 x 10 -5 mol/L Pada titik ekivalen pAg = pCl = 5,0 [Ag 2+ ] 2 [CrO 4 2- ] = 2 x 10 -12 2 x 10 -12 [CrO 4 2- ] = ————— = 0,02M (1 x 10 -5 ) 2

19

20