EQUIV Point and Acid Base Titration

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A matter of taste • Acid: from the Latin word acere, which means "sour." All acids taste sour. • Base: All bases taste bitter. They usually feel slippery, like soap (which is basic!)‏

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Arrhenius definition Only works for ‘wet’ chemistry • Arrhenius acid delivers H+(aq) when dissolved in H2O • Arrhenius base delivers OH-(aq) when dissolved in H2O • Amphoteric substance has both acid & base properties H2O(l) → H+(aq) + OH-(aq)‏

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The process of titration is used to calculate the concentration of a solution by neutralising it and doing some maths

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Titration Terminology • Titrant: Solution dispensed from buret – standardized using a primary standard (high purity material of known composition)‏ • Equivalence Point – Moles of titrant equal to unknown in solution being titrated • End Point: Perceived end of the titration, may or may not coincide with Equivalence Point – Depends on indicator choice or other method used for

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Strong Acid-Strong Base Titrations The equivalence point of the titration is the point at which exactly enough titrant has been added to react with all of the substance being titrated with no titrant left over. In other words, at the equivalence point, the number of moles of titrant added so far corresponds exactly to the number of moles of substance being titrated.

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Choosing an indicator In the strong acid titration, both indicators begin to change colour at the equivalence point (50 mL of base) so both work equally well. In the weak acid titration, thymol blue changes colour at the equivalence point, but methyl red begins to change colour after only 15mL of base are added, which is far from the equivalence point, illustrating the importance of choosing an appropriate indicator. http://www.ausetute.com.au/indicata.html

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Calculations 1. Write the balanced chemical equation for the reaction 2. Extract all the relevant information from the question 3. Check that data for consistency, for example, concentrations are usually given in M or mol L-1 but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multiply the volume in mL x 10-3 4. Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(M) : n = M x V 5. From the balanced chemical equation find the mole ratio known reactant : unknown reactant 6. Use the mole ratio to calculate the moles of the unknown reactant 7. From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(M): M = n ÷ V

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Examples 30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid 1. Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)‏ 2. Extract the relevant information from the question: NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ? 3. Check the data for consistency NaOH V = 30 x 10-3L , M = 0.10M HCl V = 25.0 x 10-3L, M = ? 4. Calculate moles NaOH n(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles 5. From the balanced chemical equation find the mole ratio NaOH:HCl 1:1 6. Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point 7. Calculate concentration of HCl: M = n ÷ V n = 3 x 10-3 mol, V = 25.0 x 10-3L M(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12M or 0.12 mol L-1

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3. http://www.youtube.com/watchv=Y-5QJIr7Xm4&feature=related 4. http://www.youtube.com/watch?v=g8jdCWC10vQ&feature=related 2. http://www.youtube.com/watch?v=6Y4Y-__ME60&NR=1 1. http://www.youtube.com/watch?v=RF40cI2O16U&feature=related

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THE END