资源描述
Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,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+).,CH,3,COOH CH,3,COO,-,+H,+,Acidic buffer 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,CH,3,COOH and CH,3,COONa,Alkaline buffer solutions,An alkaline buffer solution has 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.NH,3,and NH,4,Cl,How do buffers work?,It has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it-otherwise the pH will change.,Below is a weak acid,if we added CH,3,COONa,its like adding CH,3,COO,-,CH,3,COOH CH,3,COO,-,+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 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 buffer 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(HC,3,H,5,O,3,)and 0.10 M of sodium lactate?For lactic 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.HC,3,H,5,O,3,C,3,H,5,O,3,-,+H,+,HC,3,H,5,O,3,C,3,H,5,O,3,-,H,+,0.12 M,0.10 M,0,-x,+x,+x,0.12 x,0.10+x,x,Ka=1.4 x 10,-4,=,C,3,H,5,O,3,H,+,HC,3,H,5,O,3,Because of the small Ka,we expect x to be small relative to 0.12 and 0.10 M,so we can simplify 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:,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 HC,2,H,3,O,2,and 0.300 mol of NaC,2,H,3,O,2,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 changes),HC,2,H,3,O,2,+H,2,O,H,3,O+,C,2,H,3,O,2,-,HC,2,H,3,O,2,H,2,O,H,3,O,+,C,2,H,3,O,2,-,Initial,0.300,-,0,0.300,Change,-0.020,-,+x,+0.020,Final,0.280,-,x,0.320,Ka=(x)(0.320)/(0.280)=1.8 x 10,-5,so x=H,3,O,+,=(0.280)(1.8 x 10,-5,)/0.320 =1.6 x 10,-5,pH=-log,(,1.6 x 10,-5,)=4.80,
展开阅读全文