[JagerCookies]

This is all the information we were told. I know of some formula's but im unsure how to work this out. Thanks

Provided with a standard 0.01M solution of sodium carbonate
Exactly 25.0cm3 of the approximate 2M hydrochloric acid was very carefully diluted using a bulb pipette. The meniscus was exactly on the line of 25.0cm3 on the pipette. The hydrochloric acid was transferred into a 250cm3 volumetric flask and filled to the mark with distilled water. The flask was shaken to mix the solution and the flask labelled.
A burette was rinsed with distilled water and then filled fully to the mark with the diluted hydrochloric acid.
A rinsed pipette was used to transfer 25cm3 of sodium carbonate solution to a rinsed titration (conical) flask.
A few drops of methyl orange indicator was used to turn the solution a yellow colour.
The hydrochloric acid was titrated into the sodium carbonate solution until the indicator changed, marking the end point, the solution changed into a peach colour solution.
Burette readings were recorded
The process was repeated until three concordant results within ±0.05cm3 was achieved and recorded

2HCL(aq) + Na2CO3(aq) à 2NaCl(aq) + H2O(l) + CO2(l)
Final burette reading:24cm3

Calculate molarity of hydrochloric acid and concentration in g/dm3.

[gearoid94]

Calculate the moles of sodium carbonate used. Since it reacts with HCl in a 1:2 ratio, multiply your answer by 2 to get the moles of HCl in 25cm3. For the molarity you have to have mol/dm3 so in order to get from 0.025dm3 (25cm3) to 1dm3, divide by 0.025. Then to get g/dm3 you have to think about the equation n=m/RMM, therefore m=n x RMM, so multiply your previous answer by the RMM of HCl