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Why add excess KI in iodometric titration?
excess KI
What is the effect of adding the sulfuric acid before KI in iodometric titration?
Sulphuric acid helps in maintaining pH around 3-4 which further helps in liberation of iodine upon adding KI solution.
What is the difference between iodometric titration and iodimetric titration?
teri maa ki chut sale tum btao.....
How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?
55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles. 20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles. molarity of final diluted solution=22.9/20.5=1.117M since the no. of moles of KI present in initial and final solution are same. let.V(in ml) be the final volume of diluted solution. 222.75/V=1.117 V=199.41 ml final volume =199.41 ml
What is the role of KI in the estimation of aniline?
ki is added to liberate iodine gas . this liberated iodine gas was then titrated with sodium thiosulphate to give a permanent white precipitate. this white precipitate indicates the endpoint of the titration..
Why add excess KI in iodometric titration?
excess KI
What is the effect of adding the sulfuric acid before KI in iodometric titration?
Sulphuric acid helps in maintaining pH around 3-4 which further helps in liberation of iodine upon adding KI solution.
What is the difference between iodometric titration and iodimetric titration?
teri maa ki chut sale tum btao.....
How many grams of KI are in 25.0ml of a 3.0 (mv) KI solution?
55 ml of a 4.05 M solution of KI solution contains 55*4.05=222.75 millimoles. 20.5 ml of the diluted solution contains 3.8g of KI,so no.of moles of KI=3.8/(mol.wt of KI=165.9) is 22.9 millimoles. molarity of final diluted solution=22.9/20.5=1.117M since the no. of moles of KI present in initial and final solution are same. let.V(in ml) be the final volume of diluted solution. 222.75/V=1.117 V=199.41 ml final volume =199.41 ml
To 225 mL of a 0.80M solution of KI a student adds enough water to make 1.0 L of a more dilute KI solution What is the molarity of the new solution?
0.18M
What is the molarity of 67.94 mL aqueous solution containing 2.822 g of KI?
Molarity = moles of solute/Liters of solution. get moles KI 2.822 grams KI (1 mole KI/166 grams) = 0.017 moles KI ( 67.94 ml = 0.06794 Liters ) Molarity = 0.017 moles KI/0.06794 Liters = 0.2502 M KI
What is the role of KI in the estimation of aniline?
ki is added to liberate iodine gas . this liberated iodine gas was then titrated with sodium thiosulphate to give a permanent white precipitate. this white precipitate indicates the endpoint of the titration..
What is the molarity of a solution prepared by dissolving 2.41 g of potassium iodide KI in 100 mL of water?
Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================
How do you prepare iodine solution?
dissolve iodine crystals in a saturated KI(potassium iodide) solution
Why potassium thiocyanate is added in the titration of sodium thiosulphate with copper?
On addition of the KI to your copper (II) solution, you formed Copper (I) iodine solid and produced the tri-iodide ion. It is the tri-iodide ion that you are titrating with the sodium thiosulfate. The tri-iodine ion is what itercalates into the starch molecules to form the dark blue color you are using as an end point in the titration. Some the the tri-iodide ion formed will adsorb to the surface of the solid copper (I) iodine formed. This must be desorbed for a complete titration. The addition of the potassium thiocyanate, displaces the adsorbed tri-iodine ion, and liberates it for titration.
What to do for starch testing?
Use 'I2 + KI ' solution: 0.1% 'Iodine-Iodide' turns purple blue with starch, very sensitive test!
What mass of KI is present in 276 mL of a 0.50 M solution?
V*c*M = 276(mL)*0.001(L/mL)*0.50(mol KI/L)*166.0(g KI/mol KI) = 22.9 g KI
