[Tipsy]

I'm ready to pull my hair out 

These are the first two of my questions in my assignment:

Determine the weight of potassium hydroxide required to prepare 40ml of an 0.02M solution.

This is how I read it:

I need to work out how many grams I need out of 1 mole of KOH. 1 mole of KOH = 56.11gm. Therefore, a 1 Molar solution would require 56.11gm. However, I want a lesser concentration and volume so I should end up with a smaller number of grams, i.e a figure under 56.11gm.

I know all units must be the same in calculations, i.e I can't calculate using kg's and g's or litres and mls so need to change them to the same unit.

I know one mole of a substance is what is required to make a one litre molar solution

I know the formulas concentration = moles/volume (c = n/v) and C1V1=C2V2. However, I do not quite understand where you apply them.

My next question is "What weight of NaOH is required to prepare
a) 2 litres of 0.05M solution
b) 1 litre of 0.1M solution

A) is much like my first question

B) the answer is 4gm. I know this because what it is asking (I talk myself through these!) is that I want 1/10th (0.01M) of a 1M solution and 1/10th of 40gm is 4gm so ratio is what I used to determine this. However, I don't know how to calculate it and have to show my workings.

So far i have managed to smoke too many ciggies and use a large quanity of paper, ink, time battery power and liquid paper and still am no nearer to my answer. 

Please can someone explain it to me, don't just give me the answer.

Many thanks.

BTW I am 40 and am a nurse and have returned to study Lab technology and haven't done chemistry since I did my high school exams in 1985 where i managed to get a huge 18% final mark...oh the shame 

[sjb]

I need to work out how many grams I need out of 1 mole of KOH. 1 mole of KOH = 56.11gm. Therefore, a 1 Molar solution would require 56.11gm.

And what else? What is the definition of molarity?

However, I want a lesser concentration and volume so I should end up with a smaller number of grams, i.e a figure under 56.11gm.

Correct

I know the formulas concentration = moles/volume (c = n/v) and C1V1=C2V2. However, I do not quite understand where you apply them.

C1V1=C2V2 is not really relevant here, in my opinion, just concentrate on
concentration = moles/volume (with applicable units) for now

My next question is "What weight of NaOH is required to prepare
a) 2 litres of 0.05M solution
b) 1 litre of 0.1M solution

A) is much like my first question

B) the answer is 4gm. I know this because what it is asking (I talk myself through these!) is that I want 1/10th (0.01M) of a 1M solution and 1/10th of 40gm is 4gm so ratio is what I used to determine this. However, I don't know how to calculate it and have to show my workings.

In some ways, you have given your workings for B here. Consider the notes at http://en.wikipedia.org/w/index.php?title=Concentration&oldid=211619211

[Astrokel]

Just to add on only 

n = cv

n = mass/Mr

n = no.of particles/Avogadro's constant

n = volume of gas/molar volume

as long as you are able to get the moles, it's easy to convert to the one you want 

[Tipsy]

Determine the weight of potassium hydroxide required to prepare 40ml of an 0.02M solution

C=n/v

0.02= n/40

n= 40/0.02

n=0.0005 m

so 56.11 x 0.0005 = 0.028gm KOH

Is that correct? It seems an awful small figure but then it's small volume and concentration

[Borek]

0.02= n/40

n= 40/0.02

No. Check your math.

Honestly nothing is OK. Where did you get 0.0005 from? No matter how I try, I can't get to this number.

[Tipsy]

Do you know, I honestly can't remember where I got it from 

I just don't understand it. Sorry. I've got 15 questions to do, so far I haven't even got one done.

[enahs]

Concentration (in molarity):     # of mols              
                                      volume of solution (in L)


molarity * volume = # of mols


     # of mols               * volume  of solution (in L) = # of mols
volume of solution (in L)

How many mols of KOH are needed? What is the mass of that?

[Tipsy]

molarity x volume = no. moles

0.02 x 0.04 = 0.0008 moles.

56.11 x 0.0008 = 0.044888gm

= 0.045 gm of KOH needed.

I know that's correct because I knew the answer all along, I just didn't know how to get there!

So, it looks like I got my c=n/v wrong when changing it around to get the missing figure. If that is where I went wrong then I guess i just have to keep practising the interchanging.

Thankyou so much.

[enahs]

Always put your units and cancel them, and it will tell you if you did your math wrong!