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How Much Acid there is in a Solution

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How Much Acid there is in a Solution Analysing Evidence and drawing conclusions This is the reaction between the sodium carbonate solution and the sulphuric acid: H2SO4 (aq) + Na2CO3 (aq) Na2SO4 (aq) + CO2 (g) + H2O (l) From this equation it is clear to see that one mole of sulphuric acid (H2SO4) reacts with one mole of sodium carbonate (Na2CO3), to produce a neutral salt sodium sulphate (Na2SO4), carbon dioxide (CO2) and water (H2O). Firstly, I am going to calculate the concentration of the sodium carbonate that was used. This is done by using the formulas: Number of moles (in mol) = mass (g) � relative formula mass (g) So the concentration is equal to: concentration (in mol dm-3) = Number of moles (in mol) � volume (in dm3) ...read more.


volume (in dm3) = 0.025 0.25 = 0.100 mol dm-3 The concentration of the sodium carbonate solution that I made up is 0.1 mol dm-3 . Concentration of sulphuric acid: As I know from the equation that I stated at the start of the analysis one mole of sodium carbonate reacts with one mole of sulphuric acid, the ration being 1:1. This is the point at which neutralization occurs and the solution became clear. I can use the same equation for calculating the concentration of the acid solution, as I did for the sodium carbonate soluiton. The only difference is that not all of the sodium carbonate was used for each titration. ...read more.


Therefore, 19.7cm3 of sulphuric acid with a concentration of 0.13 mol dm-3 , neutralises 25cm3 of sodium carbonate solution with a concentration of 0.1 mol dm-3 . I think my results are accurate and reliable as all my results for the titre were close to each other, only having a difference of 0.1 cm3. My result shows that the concentration of sulphuric acid is 0.13 mol dm-3 . This is not an exact value of the concentration of the acid, but it does agree with the statement at the start of the experiment, that it is between 0.05 and 0.15mol dm-3. The main reason for my results not being completely accurate is due to human error. This will be explained further in my evaluation. ...read more.

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