calculating concentration of acid ; include all steps in working out the answer?

Question

a standard solution was prepared by dissolving 2.6061g of anhydrous sodium carbonate in deionized water, and the solution dilated to 250cm3. a 25cm3 portion of this solution was titrated against hydrochloric acid, using a suitable indicator. the endpoint was reached after 18.7cm3 of acid had been added. calculate the concentration of the acid.

the answer given is actually 0.263 mol dm-3, but i just dont know how to get to that, so can you show all the steps of your working out of how to get to that answer. thanks

Answer 1

First, write down the equation:-

Na2CO3(aq) + 2HCl(aq) = 2NaCl(aq) + CO2(g) + H2O(l)

NB 1 mole of Na2CO3 neutralises 2 moles of HCl.

Next find the Mr (Na2CO3):-
2 x Na = 2 x 23 = 46
1 x C = 1 x 12 = 12
3 x O = 3 x 16 = 48
Add = 106 (Mr - Na2CO3)
Next find the moles of Na2CO3
Moles = mass/g divided by Mr
= 2.6061/ 106 = 0.02459 (4sf) (moles Na2CO3)
and this is dissolved into 250 cm^3 (quarter litre).
So into 25 cm^3/250cm^3 ( 1/10th) of the dissolved Na2CO3 moles is present.
So in 25cm^3 there are 0.002459 moles (Na2CO3) present.
From the equation above:-
0.002459 is equivalent to 1
therefore 0.002458 x 2 is equivalent to 2. = 0.004918 moles (HCl)
0.004918 moles(HCl) is contained in 18.7 cm^3.

Using moles = conc'n x volume / 1000 cm^3

By algebraic manipulation
conc.n = moles x 1000 cm^3/volume

[conc] = 0.004918moles x 1000cm^3 / 18.7cm^3

[HCl] = 0.26299 moles dm^-3

[HCl] = 0.263 (3sf)

Answer 2

Molecular wt of anhydrous sodium carbonate = 106
moles of anhydrous sodium carbonate =2.6061/106 = 0.02458
Equivalents of anhydrous sodium carbonate =0.02458*2 = 0.04917

V1 =250 cc = 0.25L
Normality = 0.04917/0.25 =0.19668 N
V2= 18.7 cc
Apply N1*V1 = N2* V2 [ conservation of equivalents]

0.19668 * 25 = N2 * 18.7
N2 = 0.26295 N
concentration of the acid = 0.26295 N ~0.263 N
=0.263 mols/dm^3

Answer 3

Na2CO3 + 2HCl -> 2NaCl + H2O + CO2

moles of Na2CO3 = 2.6061/106 = 0.0246

since one tenth of the solution was used, the amount of Na2CO3 that reacted = 0.00246

from the equation we know that twice a many moles of HCl were involved in the reaction

moles of HCl = 0.00492

concentration of HCl = 0.00492/0.0187 = 0.263 mol dm-3

Answer 4

I leave chemistry at work, because I'm good at it, and suggest you won't get any better at it until you do your own homework.

All the answers have been laid out before you, and in truth, you should be able to work this out for yourself.

Answer 5

You people scare me.