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Titrations & Buffer solutions

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Presentation on theme: "Titrations & Buffer solutions"— Presentation transcript:

1 Titrations & Buffer solutions

2 Acid-Base Titrations Method based on measurement of volume
If concentration of acid is know, the amount of base can be found If concentration of base is known, then we can determine the amount of acid The titrant is typically a strong acid or base Sample species can be either a strong or weak acid or base.

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5 Titration Curves There are 4 types of titrations for simple acids or bases: Strong acid vs strong base Weak acid vs Strong base Strong base vs. strong acid Weak base vs. strong acid

6 Strong acid with Strong Base
Do the stoichiometry. There is no equilibrium . They both dissociate completely. Weak acid with Strong base There is an equilibrium. Do stoichiometry. Then do equilibrium.

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10 Construct the titration curve for the titration of 100mL of 0
Construct the titration curve for the titration of 100mL of mol/L HCl with M NaOH. 1. At the start After adding mL base 3. after adding At equivalence mL point 5. after adding 10mL more

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12 Indicators Weak acids that change colour when they become bases
weak acid written HIn HIn H+ + In- clear red Equilibrium is controlled by pH End point - when the indicator changes color.

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17 The Common Ion Effect When the salt with the anion of a weak acid is added to that acid, It reverses the dissociation of the acid. Lowers the percent dissociation of the acid. The same principle applies to salts with the cation of a weak base.

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30 This is called the Henderson-Hasselbach equation
pH = pKa + log([A-]/[HA]) pH = pKa + log(base/acid) Calculate the pH of the following mixtures M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)

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32 Buffered solutions-summary
A solution that resists a change in pH. Either a weak acid and its salt or a weak base and its salt. We can make a buffer of any pH by varying the concentrations of these solutions. Same calculations as before. Calculate the pH of a solution that is .50 M HAc and .25 M NaAc (Ka = 1.8 x 10-5)


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