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sulphuric acid is dibasic

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Introduction

Demonstrating that sulphuric is dibasic: Aim: The aim of my plan is to demonstrate that sulphuric acid is dibasic. Introduction: Acid is a substance that reacts with a base to form a salt and water similarly a base is a substance that reacts with an acid to form a salt and water. "Lewis (a scientist) states that an acid is a compound that accepts a pair of electrons from a base and a base is a molecule or an ion that donates a pair of electrons to an acid. Acids taste sour and are corrosive and bases are slippery". ? Acid + Base � Salt + Water Acid + Metal � Metal Salt + Hydrogen gas Prediction: I predict that sulphuric acid, H2SO4 is dibasic because it forms two hydrogen (H+) ions in an aqueous solution. It is the molecule of sulphuric acid which changes to sulphate ions (SO42-) and hydrogen ions (H+) when dissolved in water. However, hydrochloric acid is monobasic as it only forms one hydrogen (H+) ion in an aqueous solution. To prove this, I am going to carry out two different experiments that involve a titration and collection of a gas. I assume that it should require me half a volume of sulphuric acid to neutralise sodium hydroxide, NaOH than HCl in titration. ...read more.

Middle

11. If the colour of solution is going lighter, twist the tap so that only drops of acid flow into the conical flask. 12. Keep adding the drops until one last drop changes the colour of the solution to colourless. 13. Close the tap with that last drop and record the reading of the acid that was used in a table. 14. Repeat the same procedure 3 more times and find the average volume of acid used. 15. Carry out exactly the same procedure from step 1 to 13 but using HCl instead of H2SO4. 16. See the figure below for how to set up the apparatus. Pipette Solution Sodium hydroxide (NaOH) 25cm3 moldm-3 Burette Solution Acids (H2SO4/HCl) 25cm3 1.00moldm-3 Indicator phenolphthalein Burette readings Trial 1 2 3 4 Final 25.90 25.80 25.60 25.40 25.40 Initial 0 0 0 0 0 Volume used (titre)/ cm3 25.90 25.80 25.60 25.40 25.40 Mean titre/ cm3 (25.80+25.60+25.40+25.40)/4 =25.60 Since I have got two reading similar, I can stop titration. Titration: Equations H2SO4 (aq) + 2NaOH (aq) �Na2 SO4 (aq) + 2H2O (l) Concentration (H2SO4) =1.00moldm-3 Volume (H2SO4) = 25.60 cm3 no. of moles of (H2SO4) = 1.00 x (25.60/1000) = 0.025mol no. of moles of (NaOH) = 0.025 x 2 = 0.050 mol From this equation we can see that molar ratio of H2SO4: NaOH is 1:2.So, 1 mole of H2SO4 requires two moles of NaOH and produces two moles of water in titration. ...read more.

Conclusion

1/ 24dm3 =4.00x10-3 /x x = 0.096 dm3 x = 96.00cm3 Volume of H2 = 96.00cm3 Mg(S) + 2HCl (aq) � MgCl2 (aq) + H2 (g) Concentration of (HCl) = 1.00moldm-3 Volume of (HCl) = 4.00 cm3 No. of moles of ( HCl) = 1.00 x (4.00/1000) = 4.00x10-3 mole No. of moles of (H2) = 4.00x10-3 /2 = 2.00x10-3mole Since 1 mole of a gas occupies 24dm3 of space. Therefore, 1mole: 24dm3= 4.00x10-3 moles: x dm3 (where x is the volume of hydrogen gas produced.) 1/ 24dm3 =2.00x10-3 /x x = 0.048 dm3 x = 48.00cm3 Volume of H2 = 48.00cm3 1 mole of H2SO4 gives 96.00cm3of H2 gas where as, 1 mole of HCl produces only half a mole of H2 gas (48.00cm3). Consequently, I have proven that H2SO4 is dibasic. Analysis: 1. Titration gives an accurate reading because the apparatus used such as pipette and burette are accurately calibrated and so can measure up to 0.10 cm3 than the collection of gas using a measuring cylinder. 2. Titration is more reliable as it is repeated several times and an average volume of titres is found. 3. Since I am using an indicator in titration, I can determine the end point whereas there isn't an end point for the gas evolved. 4. The results of the gas collection may be inaccurate because some of the gas may have escaped while the metal is added into the acid. ...read more.

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