[blurry]

In my experiment, I added 1.31g of MCO3 (where M is the unknown alkaline metal) to 50.0cm^3 of 1.0M HCl inside a 250cm^3 volumetric flask. after the effervescence has stopped, distilled water was added to fill up the volumetric flask up to the volume graduation mark. I shall call this solution "A"
i then titrated the solution A with 0.1M NaOH, and I got an average titre of 27.55 cm^3.
i started the calculations but then i got slightly stuck.

first, i founf the amount of NaOH that reacted with the excess acid.
to do this, i did the mean titre/1000 * concentration of NaOH.
which in this case is 27.55/1000 * 0.1 = 0.002755 mol of NaOH reacted with excess acid
and since i know the reaction here was:
HCl (aq) + NaOH (aq) --> NaCl (aq) + H2O (l)
i can see that the molar ratio of HCl : NaOH is 1:1, therefore the same amount of HCl is left unreacted from the MCO3 as the amount of NaOH reacted with the excess HCl. in other words, there is 0.002755 mol of HCl that was unreacted with the MCO3

and since i know that the mass of HCl at the beginning was 50g and concentration of the HCl, i did: 50/1000 * 1 = 0.5 mol
so amount of HCl that reacted with the MCO3 is 0.5 - 0.002755 = 0.047245 mol

and i know the equation here is MCO3 + 2HCl --> MCl2 + H2O + CO2
so i can see that the mole ratio of HCl : MCO3 is 2 : 1
Amt of MCO3 present = 0.047245/2 = 0.0236225 mol
If 0.0236225 mol of MCO3 weighs 1.31g, 1 mol weighs 1.31/0.0236225= 55.45g

my question is, how do i factor in the fact that i diluted the solution in the volumetric flask during the calculations? do i even need to factor it in?

are my calculations even right?? because i know that one mole of CO3 has a Mr of 60.
oh no. now im really confused.

[Borek]

50.0cm^3 of 1.0M HCl

i know that the mass of HCl at the beginning was 50g

Was it?

i then titrated the solution A with 0.1M NaOH, and I got an average titre of 27.55 cm^3.

Average - so you titrated several times. Where did you took other samples for titration from? Have you started with several samples of solid, or have you taken separate aliquots of the same solution A for titration?

[blurry]

well, i assumed that 1 cm^3 = 1 g.

and i did 2 titration, the first was 27.6 and the next was 27.5. for both titrations, i titrated the 0.1M NaOH against 25cm^3 of the solution A. i hope that clears it up a little.

[Borek]

well, i assumed that 1 cm^3 = 1 g.

So you got 50 grams of solution, not 50 grams of HCl. Mass of the solution is irrelevant, it is number of moles of HCl that counts.

and i did 2 titration, the first was 27.6 and the next was 27.5. for both titrations, i titrated the 0.1M NaOH against 25cm^3 of the solution A. i hope that clears it up a little.

25 mL of solution A, so one tenth of the solution A. What voulme of NaOH would be needed for 250 mL of solution A?

[blurry]

So you got 50 grams of solution, not 50 grams of HCl. Mass of the solution is irrelevant, it is number of moles of HCl that counts.

i'm not really sure what you're getting at here... i know this is just me being really dim, but i thought in order to find the number of moles i would have to do volume/1000 * concentration = moles? or am i really just not making any sense here?

25 mL of solution A, so one tenth of the solution A. What voulme of NaOH would be needed for 250 mL of solution A?

ah. so since 0.002755 mol of NaOH reacted with 25cm^3 of solution A, 0.02755 would have been needed to react with the total solution of A?