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Is sulphuric acid dibasic?

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Introduction

Chemistry Planning Exercise Aim In this plan I shall demonstrate that sulphuric acid is dibasic. In order to do this affectively I will plan experiment using scientific techniques. The experiment will involve titration. I will then validate these experiments using calculations and appropriate chemistry knowledge. The reactants I have selected that will neutralise sulphuric acid are shown below in the equations: Reactants used in Titration and products obtained: H2SO4 (aq) + 2NaOH (aq) --> Na2SO4 (aq) + 2H2O (l) Background knowledge The acid (sulphuric acid) is identified by being paired with a hydrogen ion and a base (sodium hydroxide) is identified by being paired with a hydroxide group. Sodium Hydroxide neutralises the sulphuric acid to form salt and water. Sulphuric acid A dibasic acid has two hydrogen atoms in its molecule which can be ionised. Sulphuric acid is dibasic acid, because it contains two hydrogen atoms which ionise in aqueous solution to become 2H+ ions. The equation below shows an ionic equation for sulphuric acid. H2SO4 + 2H+ --> SO42- Sodium Hydroxide Sodium Hydroxide is used as a base. It is completely ionic, containing sodium ions and hydroxide ions. The hydroxide ions make sodium hydroxide a strong base which reacts with acid to form water and salt. ...read more.

Middle

Do not try to induce vomiting. Call for immediate medical help. Procedure Preparing a standard solution of sodium hydroxide: You measure accurately a sample of sodium hydroxide and use it to make a solution of concentration of 0.2M. This solution will be used to determine the volume of a solution of sulphuric acid. Procedure 1 1. Measure 50cm3 of 0.4moldm-3 of solution of sodium hydroxide into a measuring cylinder. 2. Transfers the measured sodium hydroxide solution into the volumetric flask through the filter funnel. Rinse the measuring cylinder well, making sure all liquid goes into the volumetric flask. 3. Add water until the level is 1cm of the mark on the neck of the flask. Insert the stopper and shake to mix the content. 4. Using the drooping pipette, add enough water to bring the bottom of the meniscus to the mark, as in the diagram. Insert stopper and shake thoroughly ten times to ensure complete mixing. Simply inverting the flask once or twice does not mix the contents properly and may result in a fault. 5. Label the flask with sodium hydroxide (NaOH). Procedure 2 In procedure 1 you made a standard solution of sodium hydroxide up to 100cm3. ...read more.

Conclusion

Note that this does not introduce a fourth figure; it merely makes the third figure more reliable. Suitable quantities to use in both the experiment In this experiment I have decide to use 50cm3 of 0.4M of sodium hydroxide which will be made up to 100cm3. The standard solution of sodium hydroxide will have 0.2M. 50cm3 of NaOH of 0.2M. Moles = concentration x volume = 0.4x50 = 0.2 1000 1000 Specimen calculation Titration data: Conical flask reagent Sodium hydroxide 0.2mol dm3 25cm3 Burette reagent Sulphuric acid 1.00mol dm3 Indicator Trial run Run 1 Run2 Burette reading Final 28.0 25.5 24.5 Initial 0.00 0.00 0.00 Volume used (titre)/cm3 28.0 25.5 24.5 Mean titre/cm3 25.0 Step 1: Amount of NaOH = concentration x volume Amount= 0.2moldm3 x 25.0 = 0.005mol 1000 Step 2: Amount of H2SO4 = concentration x volume Amount= 1.00moldm3 x 25.0 = 0.025 1000 Step 3: Ratio of H2SO4 to NaOH 0.0025: 0.005 1 : 2 Ratio of H2SO4 to NaOH is 1:2 2 mol of NaOH reacts with 1 mol of H2SO4 so the equation is: H2SO4 (aq) + 2NaOH (aq) --> Na2SO4 (aq) + 2H2O (l) When this reaction occurs 1 mole of sulphuric acid releases 2 moles of hydrogen ions, hence proving that sulphuric acid is dibasic. ?? ?? ?? ?? Candidate name: Manal Javed Candidate number : Centre number: ...read more.

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