SCHA266WeakAcidLab20073 - Determination of the First...

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Determination of the First Dissociation Constant of Weak Acids SCHA-266 Experiment 1 Apparatus : pH meter with electrode sodium hydroxide solution, about 0.1 M (standardized) unknown solution of a weak acid (100mL of approx. solution) 25-mL pipet buret, buret clamp, and ring stand 150-mL beaker 250-mL beakers (3) standard buffer solutions, pH 4, 7 and 10 General Theory According to the Brønsted-Lowry acid-base theory, the strength of an acid is related to its ability to donate protons. All acid-base reactions are then competitions between bases of various strengths for these protons. For example, the strong acid HCl reacts with water according to Equation [1]: HCl ( aq ) + H 2 O ( l ) → H 3 O + ( aq ) + Cl - ( aq ) [1] This is a strong acid and is completely dissociated (in other words, 100 percent dissociated) in dilute aqueous solution. Consequently, the [H 3 O + ] concentration of a 0.1 M HCl solution is 0.1 M . Thus HCl is a stronger acid than water and completely donates a proton to water to form H 3 O + . By contrast, acetic acid, HC 2 H 3 O 2 (abbreviated HOAc), is a weak acid and is only slightly dissociated, as shown in Equation [2]: H 2 O( l ) + HOAc( aq ) H 3 O + ( aq ) + OAc - ( aq ) [2] Its acid dissociation constant, as shown by Equation [3], is therefore small: K a = ] HOAc [ ] OAc ][ O H [ 3 - + = 1.8 x 10 -5 [3] Acetic acid only partially dissociates in aqueous solution, and an appreciable quantity of undissociated acetic acid remains in solution. For the general weak acid HA the dissociation reaction and dissociation expression are: HA( aq ) + H 2 O( l ) H 3 O + ( aq ) + A - ( aq ) [4] K a = ] HA [ ] A ][ O H [ 3 - + [5] SCHA-266 Page 1 Weak Monoprotic Acid
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Recall that pH is defined as: –log [H 3 O + ] = pH [6] Solving Equation [5] for [H 3 O + ] and substituting this quantity into Equation [6] yields: [H 3 O + ] = K a ] A [ ] HA [ - [7] –log [H 3 O + ] = –log K a – log ] A [ ] HA [ - [8] pH = p K a – log ] A [ ] HA [ - [9] where p K a = –log K a If we titrate the weak acid HA with a base, there will be a point in the titration at which the number of equivalents of base is just one-half the number of equivalents of acid present (at ½ V e ). This is the point at which 50 percent of the acid has been titrated to produce A - and 50 percent remains as HA. At this point [HA] = [A - ], the ratio [HA]/[A - ] = 1, and log [HA]/[A - ] = 0. Hence, at this point in a titration, that is, at one-half the equivalence point, Equation [9] becomes pH = p K a @ ½ V e [10]
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