Mental Health Documentation 2020: Coding for Groups & Documenting Services
Stay updated with the latest information on coding for mental health group services in 2020. Learn about the new group code 550, important points to remember in documenting group interventions, how to determine the number of clients in a group, and the necessity of unique group descriptions. Ensure proper documentation to accurately reflect your group interventions and address clients' mental health needs.
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Oxidation Reduction Titration These include all reactions involving transition of electrons between the reactants. Oxidation: M M+n + n e n = 1, 2, 3 . Reducing agent Example: Fe+2 Fe+3 + 1e Cu+1 Cu+2 + 1e Reduction:M+n + ne M Oxidizing agent Example: Mn+7 + 5e Mn+2 Ce+4 + 1e Ce+3
Preparation and standardization of 0.01N KMnO4 solution by standard sodium oxalate (Na2C2O4)
KMnO4 is an oxidizing agent, and the oxidation can be occurring in acidic and neutral medium as the following: 1- In acidic medium KMnO4 is reduced in acid solution and the color changes from purple to colorless. MnO4- + 8 H+ + 5e Mn+2 + 4 H2O Purple Colorless M.wt 5 158 5 Eq.wt KMnO4 = = 31.6 g/mol =
2- In neutral medium MnO4- + 2 H2O + 3e MnO2 + 4OH- Purple Brown ppt. M.wt 3 158 3 Eq.wt KMnO4 = = 52.66 g/mol = In acidic medium, the activity of KMnO4 is high and the colorless Mn+2 ion was formed, while a brown precipitate of MnO2 was formed in the neutral medium, therefore the detection of the equivalence point in acidic medium was easier.
KMnO4 is not a primary standard substance, i.e. it is not possible to prepare a standard solution of KMnO4 by accurate weighing, therefore, the prepared solution of KMnO4 must be standardized before use; because it is difficult to obtain KMnO4 in pure state and free from MnO2. There are several primary standard substances for standardizing of KMnO4 such as arsenious oxide (As2O3), K4[Fe(CN)6] and sodium oxalate (Na2C2O4), but Na2C2O4 is the most popular.
The complete reaction of Na2C2O4 in acidic medium with KMnO4 during the titration process is as the following: 80 C 2KMnO4 + 5Na2C2O4 + 8H2SO4 2MnSO4 + K2SO4 + 5Na2SO4 + 10CO2 + 8H2O Oxidation C2O4-2 2 CO2 + 2e 5 Reduction MnO4- + 8H+ + 5e Mn+2 + 4H2O 2 2 MnO4- + 5 C2O4-2 + 16 H+ 2 Mn+2 + 10 CO2 + 8 H2O This reaction is called Reduction- Oxidation Reaction (Redox).
The equivalence weight of Na2C2O4 is half (1/2) of its molecular weight: Eq.wt Na2C2O4 =M.wt 134 2 = 67 g/mol = 2 In this reaction there is no need for using of indicator in the titration; because potassium permanganate (KMnO4) behaves as self-indicator, therefore one drop of KMnO4 solution is enough to produce pink color for the solution.
The dilute H2SO4 is suitable to acidify the medium of the reaction; because there is no effect of H2SO4 on KMnO4 in dilute solution, but if HCl is used the following side reaction occur: 2MnO4- + 10Cl- + 16 H+ 2Mn+2 + 5 Cl2 + 8H2O and this lead to the consumption of some of the MnO4- in oxidation of 2Cl- Cl2+ 2e
An intermediate brown coloration is sometimes seen due to the formation of MnO2 and indicating that the permanganate ion MnO4- is not being reduced to the stage of manganous ion Mn+2. Therefore the following conditions must be observed: 1- The temperature of H2C2O4 oxalic acid solution must be kept at 80 C. If a brown color appears during titration the flask should be reheated. 2- Excess of diluteH2SO4 must be present in the flask. 3- The permanganate solution must be run slowly.
Preparation of solutions 1- Prepare approximately 0.01 N KMnO4 solution in 250 mL of distilled water. N eq.wt V (mL) Wt = eq.wt of KMnO4 =F.wt of KMnO4 1000 n 0.01 250 158 5 1 39 + 1 55 + 4 16 5 = = 1000 = 0.079 gm = 31.6 g/mol Dissolve 0.079 gm of KMnO4 in 250 mL of distilled water to obtain approximately 0.01N KMnO4 solution. Transfer the solution to a large beaker. Heat to 90 C for 1hour (or alternatively allow standing for 2-3 days), cool and filter through glass wool. Transfer the solution to a dark colored bottle.
2- Prepare 0.01 N standard Na2C2O4 solution in 250 mL of distilled water. N eq.wt V(mL) eq.wt of Na2C2O4 =F.wt of Na2C2O4 Wt = n 1000 2 23 + 2 12 + 4 16 2 134 2 = 0.01 250 = 1000 = 67 g/mol = 0.1675 gm Dissolve 0.1675 gm of Na2C2O4 in 250 mL of distilled water to obtain 0.01N Na2C2O4 solution.
Procedure 1- Pipet 5 ml of 0.01N standard Na2C2O4 solution into a conical flask. 2- Add 5mL of H2SO4 (2N), then heat the content of conical flask to about 80 C. 3- Titrate slowly with KMnO4 solution (from the burette) to prevent the formation of MnO2 until the first drop gives a pink color. Repeat the titration three (3) times. V Na2C2O4(ml) VKMnO4 (ml) 5 V1 V1 + V2 + V3 3 V average = = (Y) ml KMnO4 5 V2 5 V3
5 ml of KMnO4solution Na2C2O4(0.01N) 5 ml H2SO4(2N) 5 ml of Na2C2O4(0.01N) 5 ml H2SO4(2N) Heating The Colour changed from collorless to pink (3) (1) (2)
Calculation At the equivalence point no. of milliequivalence of Na2C2O4 no. of milliequivalence of KMnO4= (N2 V2) Na2C2O4 (N1 V1) KMnO4= NKMnO4 Y = 0.01 5 0.01 5 Y NKMnO4= NKMnO4= ( ) eq/L
