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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

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14y ago
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6mo ago

The equivalent weight of KMnO4 is the molar mass divided by its change in oxidation state, which is 5 in acidic medium and 7 in basic medium. In acidic medium, the equivalent weight is 31.6 g/equivalent, and in basic medium, it is 22.5 g/equivalent.

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14y ago

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

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12y ago

the equivalent weight of kmno4 when converted to mno2 is 158/3

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12y ago

mol wt/5

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Q: What is the Equivalent weight of KMnO4 in different medium?
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What is the ratio of equivalent weight of kmno4 in acidic and basic and neutral medium?

The ratio of equivalent weight of KMnO4 in acidic to basic medium is 1:2:5. In acidic medium, KMnO4 reacts as MnO4^- + 8H+ + 5e^- → Mn^2+ + 4H2O, requiring 5 equivalents, whereas in basic medium, it reacts as MnO4^- + 2H2O + 3e^- → MnO2 + 4OH^-, requiring 2 equivalents. In neutral medium, the equivalent weight is the same as in basic medium.


What is the equivalent weight of potassium chromate What is the equivalent weight of KMnO4 in neutral medium?

The equivalent weight of potassium chromate (K2CrO4) is equal to its molar mass divided by the change in oxidation number of the central atom upon reduction. For KMnO4 in a neutral medium, the equivalent weight is calculated similarly, taking into account the change in oxidation number of manganese from +7 to +2 upon reduction.


In manufacture of potassium permanganate by electrolysis of k2mno4 what should be the equivalent weight of kmno4 how much kmno4 will be produced by 96478 coulombs or one farady?

The equivalent weight of K2MnO4 is 158 g/mol. To calculate the produced KmnO4, we first determine the number of faradays in 96478 coulombs (1 F = 96485 C). Then we use the stoichiometry: 1 mol of K2MnO4 produces 1 mol of KMnO4. So, 158 g of K2MnO4 will produce 158 g of KMnO4.


What is the molarity of 0.1 normal KMnO4?

The molarity of 0.1 normal KMnO4 solution can be calculated by multiplying the normality by the equivalent weight of KMnO4. In this case, since KMnO4 is a monobasic salt (one equivalent per mole of KMnO4), the molarity will be 0.1 M.


WHAT IS THE MEDIUM FOR CONVERTING KMnO4 TO K2MnO4?

In order to convert KMnO4 to K2MnO4, a reducing agent is typically needed. One common method involves using a reducing agent like a metal, such as zinc or iron, to react with KMnO4 in an acidic medium to produce K2MnO4.


What is the equivalent weight of Bayer's reagent?

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..


What is the amount percentage of manganese in potassium permanganate KMnO4?

Manganese makes up about 31.65% of the molecular weight of potassium permanganate (KMnO4).


How do you prepare the 1N concentration of potassium permamganate solution?

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.


Why use h2so4 instead of hcl in titration of kmno4?

H2SO4 is typically used instead of HCl in the titration of KMnO4 because HCl can react with KMnO4 and form chlorine gas, which can interfere with the titration results. Additionally, H2SO4 provides the required acidic medium for the reaction to occur between KMnO4 and the analyte.


Why acid is added in KMnO4 titration?

Acid is added in KMnO4 titration to provide an acidic medium, which helps to stabilize the oxidizing agent MnO4-. The acidic solution also helps to prevent the premature reduction of permanganate ions and ensures a clear endpoint in the titration by facilitating the reaction with the analyte.


Chemical symbol for permanganate?

The formula for potassium permanganate is KMnO4


How to prepare a 0.2N KMnO4 solution?

Formula weight of KMnO4 is 158.04. The equivalent weight depends on the reaction involved. In acidic media it is (formula weight) /5, in neutral although this is more difficult to control it is the formula weight. Knowing which type it is usually acidic as it is easier, you will need 158.04/5 X 0.2 = 6.32g. Sulfuric acid is usually added in the titration to ensure the correct oxidation reaction is carried out. Also note KMnO4 is not considered a primary standard so you should really filter the solution to remove any MnO2 caused by oxidation of organic impurites in the water you are using, and standardise the solution by titration against say oxalic acid. K