how to prepare 0.5N alcoholic KOH if the KOH is 46.4% (conc.)?

Question

the method which i have got is using the KOH which is in powder form,but i have in liquid form,so how to make the solution?...thank you so much ...urgent...

Answer 1

Calculation:
0.5N X 56.10564 g/mol / .464 = 60.459 g (per 1 liter total volume)

Procedure:
Weigh out 60.459g of 46.4% KOH
Add to a 1000 ml volumetric flask
Fill the balance of the flask with the alcohol or other solvent needed.

Hope this helps!

Answer 2

Normality and molarity are related through the equation
N=aM
where a is the number of H+/OH- per molecule for acid/bases or number of electrons gained/lost for oxidants/reductants.

Here you have a monoprotic base so a=1 and N=M

So your problem boils down to preparing a 0.5 M solution.

Your stock solution is 46.4% meaning that it is
46.4 g of KOH per 100 mL solution =>
464 g / L and if we divide with the molecular weight of KOH we get
464/56.11 mole/L = 8.269 M

So now we use the classic equation for dilution. If you want to prepare V2 mL of 0.5M solution you need

C1V1= C2V2 =>
V1= (C2/C1)*V2 = (0.5/8.269)*V2 = 0.0605*V2 mL of the stock.

Take V1 mL of the stock, put it in an appropriate volumetric cylinder or flask, add solvent up to V2, mix and you're done.

Answer 3

The KOH is probably 46.6% weight/volume and so you just need to calculate the molarity of the solution.

46.6% means that there is 46.6g of KOH for every 100 mL of solution. So, just figure out the molarity and then go from there.

But only do this if you are sure that it is 46.6% KOH weight/volume or w/v.

Answer 4

What is the molarity of the KOH solution? Find that out and you can use c1(v1)=c2(v2)