[sarah9625]

How do I find the standardized concentration of NaOH in 250ml of distilled water?

[Sev]

What do you mean?  I wouldn't think there would be any NaOH if it has been distilled.  If you have added it after distillation, you need to know how much you added.

[sarah9625]

NaOH was in the form of a solid. 0.102g of 0.1M NaOH was added to 250ml of distilled water to make a mixture. That mixture was then placed in a pipet to be used for a titration, later. Then 25ml of water and 0.3g of KHP and 3 drops of indicator were mixed. These samples were titrated with the NaOH in the pipet.

A column was then prepared with resin, 3M HCL and distilled water so that the pH of the column matched that of the distilled water. Then the sampling started: 25ml of tap water was sent through the column three times in 3 different beakers. These three beakers were then titrated using the NaOH in the pipet.

I have the data from this lab. I am not sure how to use it to claculate the total positivly charged ions in the sample. I know that OH- moles = H+ moles, thats all.

I also need to find what the standardized concentration of NaOH is. I think it is 0.102g in 250ml NaOH, but that may be wrong.

[Sev]

NaOH was in the form of a solid. 0.102g of 0.1M NaOH was added to 250ml of distilled water to make a mixture.

Not sure what you mean by 0.1M.  I also assume you made up the 250mL solution by adding water to solid. 
Finding the standardized concentration is just a matter of stoichiometry.  Work out number of mole in 0.102g, use this to work out molarity.

[sarah9625]

for my first sample:
0.301g of KHP / 204.22g=0.00147 moles / 0.1631L= 0.009Molarity. Is this the number of positivly charged ions in the sample?

[LQ43]

Not sure what you are exactly trying to find here.
1. do you mean the concentration of NaOH that will be determined by the titration with KHP?
This is a classic way of determining the conc of an unknown NaOH solution. It seems here that you have two ways of determining NaOH conc., directly with mass of NaOH and 250mL (was this from a volumetric flask?)

NaOH was in the form of a solid. 0.102g of 0.1M NaOH was added to 250ml of distilled water to make a mixture.

Finding the standardized concentration is just a matter of stoichiometry.  Work out number of mole in 0.102g, use this to work out molarity.

or indirectly from a precise (0.301 g is better than 0.3 g) mass of KHP and its titration with NaOH. Since you did this titration, the stoichiometry should always start with a balanced equation..

2. total number of positively charged ions. This also needs a dissociation equation from all the reactants involved and you can get this from doing an ionic equation from the molecular equation from above for the KHP titration

Also from WHICH reaction are you looking for this total number from. there a number of procedures that you mention, but haven't fully described

for my first sample:
0.301g of KHP / 204.22g=0.00147 moles / 0.1631L= 0.009Molarity. Is this the number of positivly charged ions in the sample?

think about this, would Molarity, a concentration term give the total number of pos charged ions?

3. just another little thing, do you mean "buret"when you say "pipet"?

4.

NaOH was in the form of a solid. 0.102g of 0.1M NaOH was added to 250ml of distilled water to make a mixture. That mixture was then placed in a pipet to be used for a titration, later. Then 25ml of water and 0.3g of KHP and 3 drops of indicator were mixed. These samples were titrated with the NaOH in the pipet.

A column was then prepared with resin, 3M HCL and distilled water so that the pH of the column matched that of the distilled water. Then the sampling started: 25ml of tap water was sent through the column three times in 3 different beakers. These three beakers were then titrated using the NaOH in the pipet.

I have the data from this lab. I am not sure how to use it to claculate the total positivly charged ions in the sample. I know that OH- moles = H+ moles, thats all.

Again, molecular/ionic equations or dissociation equns would be helpful here for THIS reaction. Even if its a simple relationship, its always good to have this, especially for this question of total number of pos ions

[sarah9625]

so for the + charged ions I have:
Sample #1: 0.301 x 1/204.22g=0.00147moles x 2= 2.94x10^-3
Sample #2: 0.303 x 1/204.22g=0.01418 moles x2= 2.84x10^-2
Sample #3: 0.303 x 1/204.22g=0.01420 moles x2=2.84x10^-2
 
And
for the standard concentration I have:
A.  2.94x10^-3 /
B.  2.84x10^-2 /
C.  2.84x10^-2 /
DIVEIDED by WHICH AMOUNT OF NaOH, The 250ml or the 16.31ml I used to titrate the KHP or the 0.90ml I used to titrate the tap water?

[LQ43]

please take a minute to go back and look at the chemistry, I sense your frustration but we're just trying to help and some of the information given is contradictory.

standard conc of NaOH:
1. please confirm the data that you first weighed out is 0.102g NaOH dissolved in 250mL of water. Calculate molarity from this data, what is it? this is not close to 0.1M - there might be a typo here

2. KHP titration - I think this is where you get the standard conc

what is the balanced equation here?
0.301/204.44 = moles KHP = moles NaOH - WHY?
moles NaOH /volume NaOH in L (.01631L) = standard conc NaOH
this IS close to 0.1M

#1 and #2 don't match and they should.

so for the + charged ions I have:
Sample #1: 0.301 x 1/204.22g=0.00147moles x 2= 2.94x10^-3
Sample #2: 0.303 x 1/204.22g=0.01418 moles x2= 2.84x10^-2
Sample #3: 0.303 x 1/204.22g=0.01420 moles x2=2.84x10^-2
 

I think this is correct #MOLES of + ions but check your mole values in red above. To get total NUMBER, how do you convert moles to number of ions?

For the titration of the tap water, are you just looking for concentration of the acid or total # of + ions here too (not sure what useful info that would be)?

[Padfoot]

Ok, from what I can see, the experiment seems to be made up of 2 parts which are:
1. The standardisation of NaOH
2. The tap water investigation

It looks like the purpose of the first part is to determine an accurate concentration of your NaOH solution, which you need for your second part.  I agree with LQ43 that you get your standard concentration from your titration results (otherwise the titration would serve no point).  Your mass of KHP was probably much more accurate than you indicated by '0.3g'.

To calculate your standard concentration of NaOH, first calculate your amount of KHP in mol that was used in the titration.

As LQ43 stated, you need a reaction eqn now:
NaOH(aq) + KHC8H4O4(aq)→ H2O(l) + KNaC8H4O4(aq)

Using the 1:1 ratio between KHP and NaOH, you can work out the amount NaOH that was in your titre volume.
Simply divide this amount by your titre volume in litres to get you standardised concentration.

I'm not sure if I exactly follow what your doing for the next part but LQ47 gave some hints.

Hope that helps 

[LQ43]

Ok, from what I can see, the experiment seems to be made up of 2 parts which are:
1. The standardisation of NaOH

NaOH(aq) + KHC8H4O4(aq)→ H2O(l) + KNaC8H4O4(aq)

Using the 1:1 ratio between KHP and NaOH, you can work out the amount NaOH that was in your titre volume.
Simply divide this amount by your titre volume in litres to get you standardised concentration.

0.301/204.44 = moles KHP = moles NaOH - WHY?
moles NaOH /volume NaOH in L (.01631L) = standard conc NaOH


Hopefully this part is clear for you now.

2. Titration of tap water to find total pos charge. (really unclear what total pos charge means)

I am guessing the question only wants the H+ ions because you are titrating with NaOH.  In tap water there are other + ions like Na+, Ca+2, etc (depending on its source) these cannot be detected with an acid base titration like you are doing.

H+  + OH- --->  H2O
HCl + NaOH --->  NaCl  +  H2O

0.90mL NaOH ---> L NaOH x M NaOH = moles NaOH

0.90ml I used to titrate the tap water?
[/quote]

yes, moles OH- = moles H+ (from equation above)

moles H+ x Avogadros number = total number of H+ ions

I am ignoring the Na+ that will be in your titration mixture since it was NOT in the origianl tap water before the titration.,

hope this helps... (my last post on this)