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Chapter-16-Tro - Chapter 16 Aqueous ionic equilibrium...

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Chapter 16: Aqueous ionic equilibrium Buffer solutions (a). Buffer solution is a mixture of significant amounts of a weak acid and its conjugate base or a mixture of significant amounts of a weak base and its conjugate acid. (b). When a solution of weak acid is prepared, it will naturally contain some amount of its conjugate base due to acid hydrolysis. But this solution is not called a buffer because of the tiny amounts of conjugate base generated during the hydrolysis of a weak acid. (c). When a solution of weak base is prepared, it will naturally contain some amount of its conjugate acid due to base hydrolysis. But this solution is not called a buffer because of the tiny amounts of conjugate acid generated during the hydrolysis of a weak base. (d). Buffer solution resists pH change (e). When an acid is added to a buffer solution, the weak base contained in the buffer solution converts the added acid to a buffer component. (f). When a base is added to a buffer solution, the weak acid contained in the buffer solution converts the added base to a buffer component.
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Suppose that a buffer is made by mixing significant amounts of weak acetic acid (CH 3 COOH) and its conjugate base, in the form of sodium acetate (CH 3 COONa). This solution will have established the equilibrium, CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) Now if you added a strong acid such as HCl(aq), then this strong acid is converted by the base component of buffer to give more weak acid : CH 3 COO - (aq) + HCl(aq) = CH 3 COOH(aq) + Cl - (aq) and a new equilibrium between weak acid and its conjugate base will be established, but with the same equilibrium constant as before adding the strong acid. On the other hand if you added a strong base such as NaOH(aq), then this strong base is neutralized by the acid component of buffer to give more weak base : CH 3 COOH(aq) + NaOH(aq) = CH 3 COONa(aq) + H 2 O(aq) and a new equilibrium between weak base and its conjugate acid will be established, but with the same equilibrium constant as before adding the strong base. Thus a buffer solution resists pH change by converting an added acid/base into the components of buffer
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(A). What is the pH of 0.10 M CH 3 COOH( aq )? CH 3 COOH( aq ) + H 2 O( l ) H 3 O + ( aq ) + CH 3 COO - ( aq ) K a =1.8x10 -5 CH 3 COOH H 3 O + CH 3 COO - Initial 0.10 M ~0 0 Change -y +y +y At equilibrium 0.10-y y y K a = 1.8 x 10 -5 = y 2 _____ (0.10-y) y= [H 3 O + ]= 1.34 x 10 -3 pH = -log 10 1.34x10 -3 = -log 10 1.34 + 3 = 2.87 (B). Add CH 3 COOH and CH 3 COONa together so that [CH 3 COOH]=0.10 M and [CH 3 COONa]= 0.50 M. What is the pH of this solution? CH 3 COONa(aq) exists as CH 3 COO-( aq ) and Na + ( aq ) in solution …..contd Solve quadratic equation. Calculating the pH of a buffer solution:
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Buffer solutions ….contd CH 3 COOH( aq ) + H 2 O( l ) H 3 O + ( aq ) + CH 3 COO - ( aq ) K a =1.8x10 -5 CH 3 COOH H 3 O + CH 3 COO - Initial 0.10 M ~0 0.50 Change -y +y +y At equilibrium 0.10-y y 0.50+ y K a = 1.8 x 10 -5 = y(0.50+y) _______ (0.10-y) y= [H 3 O + ]= 3.6 x 10 -6 pH = -log 10 3.6x10 -6 = -log 10 3.6 + 6 = 5.44 (1).
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