you need to talk to an arithmetic magician because i have NO IDEA. :D
to prepare 1N we have to dilute 40gms of NaOH in 1 litre of water as for NaOH normality =molarity so to prepare 0.1N NaOH we have to dilute 4gms of NaOH in 1 litre of water..
0.5 N
Normality= molarityX total POSITIVE oxidation number of solute Solvent=substance present in the greatest amount Solute= all other substances in the solution +1 (1 na+) is the total POSITIVE oxidation number of NaOH 2.0X1= 2.0 N=2.0
If the solution volume remains unchanged, the normality will decrease as the NaOH will react with CO2 present in the air. Of course, if the solution volume is not held constant and if the evaporation rate is sufficient to concentrate the solution - it could also increase (effectively raising the normality of the remaining solution).
The HCL concentration is 1.2M or 1.2N
to prepare 1N we have to dilute 40gms of NaOH in 1 litre of water as for NaOH normality =molarity so to prepare 0.1N NaOH we have to dilute 4gms of NaOH in 1 litre of water..
NaOH
0.5 N
Normality= molarityX total POSITIVE oxidation number of solute Solvent=substance present in the greatest amount Solute= all other substances in the solution +1 (1 na+) is the total POSITIVE oxidation number of NaOH 2.0X1= 2.0 N=2.0
If the solution volume remains unchanged, the normality will decrease as the NaOH will react with CO2 present in the air. Of course, if the solution volume is not held constant and if the evaporation rate is sufficient to concentrate the solution - it could also increase (effectively raising the normality of the remaining solution).
ACID VALUE=Normality OF KOK or NaOH*5061/w
by bottle method
The HCL concentration is 1.2M or 1.2N
Normality of NaOH = molarity of OH- , so for pure NaOH they are equal.When kept in open air the OH- ions come in contact with the slightly (and slowly) soluble carbon dioxide (CO2), which is an ACID forming oxide, thus reacting with the strong base hydroxide:CO2 + H2O -->H2CO3immediately followed by:H2CO3 + OH- --> HCO3-and eventually followed by a second step:HCO3- + OH- --> CO32-While hydroxide is 'disappearing' from solution, it decreases 'its' normality (= concentration of ions, reactive to acids)
If you have a standard solution of an acid, like hydrochloric or sulfuric, you can perform a titration in the presence of phenolphtalein or methyl orange and calculate the solution's normality or, you can weigh a sample of a strong solid acid ( orthoiperiodic acid or even oxalic acid), titrate the acid with the hydroxide solution, again in the presence of phenolphtalein or methyl orange and calculate the concentration of NaOH. If you want to have a solution with an exact concentration, let's say 1 molar, and the actual concentration is 1,33 molar, you simply calculate how much water you need to ad in a specific quantity of solution, to dilute it to exactly 1 molar.
you can't make a primay standard solution with NaOh.as NaOh is said to be a secondary standard you have to establish the normality using some primary standard acids.
25 (mL) * 0.5 (N) = 30 (mL) * X (N) X = unknown normality of acid = (0.417 N) = 0.4 N