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Grade 10 - Acid and Base - Titrations

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Grade 10 - Acid and Base - Titrations
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Neutralization & Titrations Chapter 15 Section 3
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Neutralization & Titrations

Neutralization & TitrationsChapter 15Section 3NeutralizationWhen equal amounts of a strong acid and a strong base are mixed together, water and a salt are formed. H3O+(aq) + OH-(aq) 2H2O(l)

This is because high concentrations of [H3O+] and [OH-] cannot exist at the same time[H3O+] [OH-] = 1 x 10-14

TitrationsEquivalence point: when an acid and a base are added to each other, eventually a point will come where the solution is neutralized.The gradual addition of one solution to another to reach this equivalence point is called titrationTitration is used to help determine the concentration of an acid or base. A standard solution of known volume and concentration is used to do this.

Selecting a Suitable IndicatorAll indicators have a transition range. Selecting the correct indicator is dependent on the equivalence point of the titration. We must pick an indicator where end point and the equivalence point will be the sameFor strong acid/strong base reactions the equivalence point is pH 7However, for weak acids/bases equivalence points can be lower or highter

Calculating Concentration from Titration Datan=cVAt the equivalence point [OH-] will equal [H3O+]nOH- = nH3O+Therefore we can use the following equation to work out the unknowns concentration(cH3O+)(VH3O+) = (cOH-)(VOH-) Calculating Concentration from Titration DataA student titrates 40.00 mL of an HCl solution of unknown concentration with a 0.550 M NaOH solution. The volume of base solution needed to reach the equivalence point is 24.64 mL. What is the concentration of the HCl solution in moles per liter?

http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/crm3s5_5.swf


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