The oxidation state of 'Mn' in KMnO4 is +7 after reaction as oxidizing agent 'Mn; becomes +2 so change in oxidation number is '5' the formula mass divided by change in oxidation number is equal to equivalent mass or weight, 158/5 = 31.7
KMnO4 has a Molar mass of 158.04 g/mol
There are two possibilities:
1. KMnO4 as an oxidizer in acidic media:
MnO4- + 8 H+ + 5e- --> Mn2+ + 4H2O (gained 5 electrons from reductant)
I.e. 5 Equivalents per mole, so equivalent mass of KMnO4 = M/5 = 158.04/5 = 31.61 gram/equivalent
2. KMnO4 as an oxidizer in neutral or basic media:
MnO4- + 2H2O + 3e- --> MnO2(s) + 4OH- (gained 3 electrons)
In this case: 3 equivalents per mole, so equivalent mass of KMnO4 = M/3 = 158.04/3 = 52.68 gram/equivalent
the equivalent weight of kmno4 when converted to mno2 is 158/3
mol wt/5
Alkaline soln.of KMnO4 functions as a Bayer reagent in which it turns into MnO4. Hence change in the oxidation state = 7-4=3.Sometimes equivalent weight = Original weight / Change in the oxidation state=158/3 = 52.6Hope it helps..
acidicness number of kmno4
Potassium Hydroxide is KOH and potassium permanganate is KMnO4.
tell me about the vlance of Mn in KMnO4
To create an acidic medium or in another word made iron (II) ammonium solution more acidic as kmno4 acts as a strong reducing agent only in acid medium.
determine the eqoivalent weight of KMno4 in basic medium
31.608 g/mol
Equivalent weight of KMnO4 is equal with molar weight of KMnO4. The some is and for K2MnO4, K2MnO4 - e +OH- --------- KMnO4 + KOH In general, Equivalent weight = Molar weight / Number of electrons that take or give one molecule Equivalent weight of KMnO4 = Molar weight of KMnO4 / 1
as kmno4 acts as a strong reducing agent only in acid medium
N (normality) describes a solution that contains 1 gram equivalent weight (gEW) per liter solution. An equivalent weight is equal to the molecular weight divided by the valence (here it gets a little tricky, for acids ands bases it refers to the number of H+ or OH-, in salts it must be expressed which ion is meant unless the ratio is 1:1). In the case of KMnO4, equivalent wt is reaction specific. When KMnO4 is used in acid medium as oxidiser, 5 electrons are gained by Mn atom. So equivalent wt of KMnO4 in acid medium = Molecular wt/no.of electrons gained in redox reaction = 158/5 =31.6. So for 0.1N KMnO4 solution, you have to dissolve 3.16g KMnO4 in 1L water. (Usually a little bit excess is taken, say 3.25g, since some crystals of KMnO4 will be remained undissolved that have to be removed by filtration. So eventhough u r preparing 0.1N KMnO4 solution by accurate weighing,it is not a primary standard and u have to standardise it against a primary std such as oxalic acid or sodium oxalate. In alkaline or neutral medium, reaction of KMnO4 is different and Mn gains 3 electrons in redox reaction. So, for alkaline medium redox titrations, equivalent wt of KMnO4 will be 158/3 = 52.6. So for 0.1N KMnO4 solution in alkaline medium redox titration, dissolve 5.26g in 1L water.
Alkaline soln.of KMnO4 functions as a Bayer reagent in which it turns into MnO4. Hence change in the oxidation state = 7-4=3.Sometimes equivalent weight = Original weight / Change in the oxidation state=158/3 = 52.6Hope it helps..
for preparing 0.1 normal solution of potassium permanganate you have to disssolve 3.16 g potssium permangnate in 1L water bt in alkaline or neutral medium reactions of potassium permanganate is different and Mn gains 3 electrons in redox reaction,so far alkaline medium redox titration equivalent wt of KMnO4 will be 158\3=52.6.so far,0.1 N KMnO4 in alkaline medium redox titration dissolve 5.26 g in 1L sol.
The formula for potassium permanganate is KMnO4
% by mole. There are six elements in KMnO4 only one is K so 1/6 which is 16.67% % by mass. Molar weight of K 39grams/mole. Molar weight of KMNO4 154 grams/mole so 39/154 =25%
Because KMnO4 is an internal indicator.. No need to have any indicator.. it has distinctly different colour when it is reduced
23453 ha ha
acidicness number of kmno4