# I am having trouble calculating the moles in an experiment

Submitted by jenifer_sheppard on Mon, 09/24/2012 - 11:12

I'm trying to do my assignment and I MUST be getting the first question wrong because it's messing up all my other calculations and I really can't figure it out. I'm so frustrated!

For a standardization of an NaOH solution I used 1.74g (which is noted to be approximately 0.043mol) NaOH.

For the first titration I weighed 0.1484g potassium acid phthalate in 50mL of distilled water. I titrated that with .00804L of the NaOH solution.

For the second titration I dissolved a 1.32g TUMS (CaCO3) tablet in 100mL of HCl (M = .2004). I titrated 0.025L of that solution with 0.024L of of the NaOH solution.

My questions are:
1. determine number of moles NaOH used in each of the titrations by using the volume of the NaOH used and the concentration of NaOH. (I MUST be doing something wrong here..)
2. Determine number of moles HCl that reacted with the NaOH in the back titration. (= amount NOT neutralized by acid)
3. calculate number of moles HCl neutralized by the portion of the antacid tablet by calculating the difference between the total amount HCl added to the antacid and the amount neutralized by the NaOH in the titration
4. determine number of moles of HCl neutralized by the WHOLE tablet by accounting for the fact that only part of the solution was titrated. (I also don't understand how to do this one)

I think if I can get #1 then I'm good for #2,and 3 (4's got me stuck too)

1.74g NaOH is 1.74/30 = 0.043 moles of NaOH if Na=23, O=16 and H=1. Not sure what the volume  of this is?

NaOH reacts with potassium hydrogen phthalate (KHP) in a 1:1 mole ratio. Find the molar mass of KHP, C8H5KO4. Internet tells me it is 204.22 g/mol. So 0.1484g will be 0.1484/204.22 moles of KHP. This value will also be the number of moles of NaOH = w moles. So molarity will be number of moles/volume in Litres = w/.00804

For the second part, find the total moles of HCl added to the TUMs from volume in L x molarity = 0.1000 x 0.2004. Some of this acid reacted with the TUMs and the rest was titrated against NaOH. However, you only then used 1/4 of that solution in the titration with NaOH. You have the molarity of NaOH and the volume (0.024L). Find moles NaOH in this titration from molarity x volume in L. As HCl and NaOH react in a 1:1 ratio, the number of moles of HCl will be equal to the number of moles NaOH. Then the found answer for number of moles 'unused' acid came from 1/4 of the solution after adding tums. So the total number of moles of HCl not neutralised by the tums will be 4 times the previous answer.

So you know the number of moles of acid you had before you added the tums, and you know the number of moles of HCl that were not neutralised by the tums. The difference in these two values will be the number of moles HCl that react with the tums