how to figure out the normality/equivalent weight/equivalents of chemicals in a balanced equation (assuming necessary data given)??pls give me a few pointers...tomorrow is my practical exam involving titration and i don't have time to go through all the theory involved...
normality is molarity *equivalent.. example H2SO4 react with NaOH with chemical reaction H2SO4 + 2NaOH== Na2SO4 + 2H2O .. 1 M H2SO4 can neutralized 2 M NaOH so equivalent of H2SO4 is 2, and equivalent of NaOH is 1..
so equivalent is basically the ratio of moles in which reactants react?and what is the equivalent weight then...
No. for ions, the no. of equivalents in an ion is basically the charge on it. In the above eqn, SO4(2-) =2 equivalents of sulphate ions H+ =1 equivalent of hydrogen ion 2H+ =2 equivalents & so on..
but for the H2SO4 as whole..what would be its equivalent..
this is another example : H3AsO4 + 2KOH --> K2HAsO4 + 2H2O in that reaction, equivalent of H3AsO4 is 2 because 1mole of H3AsO4 just can neutralizing 2 moles of KOH
but i think @rajathsbhat also true. if it's just ion, equivalent is based on it's charge..
for an acid or a base, the no. of equivalents in one molecule is its acidity or basicity
so basically we are just taking the ratio of no of moles of reactants as the ratio of their chemical equivalent right..
In a balanced chemical reaction, YES
what if it is not a acid base reaction..i mean will it hold for any reaction in general..
Yes.
and equivalent weight is just the weight of 1 equivalent of any compound right..
for eg, Na2CO3 + 2HCl --> NaCl + H2O +CO2 here, 1 molecule of Na2CO3 reacts with 1 eq. of HCl therefore there are 2 eq of Na2CO3
Yes to that too
but '1 molecule of Na2CO3 reacts with 1 eq. of HCl'.... isnt a equivalent containing a really large no of molecules of order of 1 mole?
Ah sorry. I meant 1 mole of Na2CO3 reacts with 2 eqs of HCl
By the way, 1 eq of HCl=1mol of HCl
is it fixed..doesnt it depend on its reaction's stochimetric coefficient ratio which may vary with each rxn?
I think it is better to discuss the actual definitions of these terms
for an acid, the no of equivalents=the no of replaceable H ions in the molecule for a base, it is the no of replaceable OH ions in the molecule
so since HCl has 1 replaceable H ion, the no of eq of this acid in 1 mole is one
ok ,so what if there is no OH in the base..like NH3..
Now if a molecule does not have H or OH ions in it, it is reacted with HCl or KOH and the no.of moles reacting with 1 mole of HCl or KOH becomes the no.of eq in 1 mole of that molecule
actually when i tried to study it a long time ago then also i found this same problem...while sometimes it was said that chemical equivalent is a relative quantity depending on a reaction,other times it was said to be found using reference to a strong acid like HCL..if it is found using a reference acid,will the answer be same if we tried to find it in a reaction...there are compounds like S2 which have variable valancy..which one to try if a reaction involve them..
Hmm, like Fe2+ & Fe3+.... wait, I will refer my trusty chemistry book and tell tou in about 10 minutes....
yes please...i'm refering RCM and it is confusing me like anything!!
Get this. no. of equivalents of a compound in a balanced reaction=mol.mass/equivalent mass the eq. mass of a substance can vary from reaction to reaction therefore the no of equivalents of a substance varies too Now if you are aware of the ways of computing the eq. masses of acids, bases, salts, oxidizing agents & red. agents, you will be able to calculate the no.of eq. of the compound in the reaction
I think you should refer RCM to know how to calculate the eq. masses and then you'll be fine
hey, in RCM there are formulae for the eq. masses of acids, bases, salts, oxidizing agents & red. agents and ions...so basically we have to remember all these and try to fit in one of them in any given chemical equation to calculate equivalent mass of a reactant?
Yup
phew that is an aweful lot of trouble..why can things be simple and derivative like in physics..anyway,thanks a lot man..it was really helpful..
Always here to help
k then i gotta go and do those formulae..thanks again..
@iasazhary thanks you too..
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