Buffers-ednetnsca：ednetnsca缓冲器.ppt

1、Buffers,Chem 12A Mrs. Kay,Buffers help maintain a constant pH. They are able to accept small quantities of acids and bases without drastically changing their pH A buffer is composed of a weak acid molecule in equilibrium with its conjugate base and hydrogen ion (H+). CH3COOH CH3COO- + H+,Acidic buff

2、er solution:,An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt. Example: acetic acid and sodium acetate CH3COOH and CH3COONa,Alkaline buffer solutions,An alkaline buffer solution h

3、as a pH greater than 7. Alkaline buffer solutions are commonly made from a weak base and one of its salts. A frequently used example is a mixture of ammonia solution and ammonium chloride solution. NH3 and NH4Cl,How do buffers work?,It has to contain things which will remove any hydrogen ions or hyd

4、roxide ions that you might add to it - otherwise the pH will change.,Below is a weak acid, if we added CH3COONa, its like adding CH3COO- CH3COOH CH3COO- + H+ There will be a shift to the left The solution will therefore contain these important things: lots of un-ionised acetic acid; lots of acetate

5、ions from the sodium acetate; enough hydrogen ions to make the solution acidic. Other things (like water and sodium ions) which are present arent important to the argument.,Look to below website for a flash on how buffers work: ,Buffer capacity and pH,Buffer capacity is the amount of acid/base the b

6、uffer can neutralize before the pH begins to drastically change. The pH of the buffer depends on the Ka for the acid and on the relative concentrations of the acid/base of the buffer,Calculations,What is the pH of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M of sodium lactate? For lac

7、tic acid, Ka= 1.4 x 10-4 Determine the important species involved Set up ice table Set up equilibrium expression Plug in concentrations at equilibrium,1. HC3H5O3 C3H5O3 - + H+,Ka = 1.4 x 10-4 =C3H5O3 H+ HC3H5O3 Because of the small Ka, we expect x to be small relative to 0.12 and 0.10 M, so we can s

8、implify our equation to Ka = 1.4 x 10-4 =0.10 x 0.12 Solve for x, x =1.7 x 10-4,Finally, solve for the pH of the buffer pH = - log 1.7 x 10-4 = 3.77 Notice, when you put the value of x into the original equation it makes no difference for your final answer.,Addition of strong acids/bases to buffers:

9、,We assume that the strong acid or base added is consumed completely by the reaction with the buffer HX H+ + X- By adding a strong acid, the shift is to the left, therefore increasing HX and decreasing X- By adding a strong base, the OH- is used up by HX to produce X-, so X- increases and HX decreases.,Calculation,A buffer is made by adding 0.300 mol of HC2H3O2 and 0.300 mol of NaC2H3O2 to 1.0 L of water. The pH of the buffer is 4.74. Calculate the pH of this solution if 0.020 mol of NaOH is added (ignore volume chang