Acids and Bases

Chapter 18 - Additional Aspects of Acid-Base Equilibria• should be familiar with problems like those done inPractice Examples 18-3 and 18-4 (calculations similarto those in Practice Examples 18-7 and 18-8 areincluded in the lecture notes)Example 18-2A (page 714)Calculate [H 3 O + ] and [CHO 2 - ] in a solution that is 0.100 MHCHO 2 and 0.150 M NaCHO 2 .- this is a buffer of formic acid and sodium formateHCHO 2 W CHO 2-+ H +Initial 0.100 M 0.150 M 0Change -x +x +xEquilibrium 0.100 - x 0.150 + x xK a = [CHO 2 - ][H + ]/[HCHO 2 ] = 1.8 x 10 -4{(0.150 + x)(x)}/(0.100 - x) = 1.8 x 10 -4- assume x

Example 18-5 (page 719)What mass of sodium acetate must be dissolved in 300.0 mLof 0.25 M acetic acid to produce a solution with a pH of 5.09?(Assume that the solution volume remains constant at300.0 mL.)CH 3 COOH(aq) W CH 3 COO - (aq) + H + (aq) K a = 1.8 X 10 -5pH = 5.09 = -log[H + ]log[H + ] = -5.09[H + ] = 10 -5.09 = 8.128 X 10 -6 M- the amount of H + is small and therefore the concentration ofacetic acid at equilibrium will be essentially the same as theinitial concentrationK a = [CH 3 COO - ][H + ]/[CH 3 COOH] = 1.8 X 10 -5[CH 3 COO - ](8.128 X 10 -6 )/(0.25) = 1.8 X 10 -5[CH 3 COO - ] = 0.554 Mmass of sodium acetate required= (moles of sodium acetate)X(MW of sodium acetate)= {[CH 3 COO - ](volume of solution)}X(MW of sodium acetate)= {(0.554 M)(0.300 L)}X(82.034 g mol -1 )= 14 g

Example 18-6A (page 723)A 1.00 L volume of a buffer is made with concentrations of0.350 sodium formate and 0.550 M formic acid.HCHO 2 W CHO 2 - + H + K a = [CHO 2 - ][H + ]/[HCHO 2 ] = 1.8 x 10 -4a) What is the initial pH?- since this is a buffer we can assume that the concentrationsof formic acid and the formate ion at equilibrium areessentially the same as the initial concentrations[H + ] = K a X [HCHO 2 ]/[CHO 2 - ][H + ] = 1.8 x 10 -4 X (0.550)/(0.350) = 2.839 x 10 -4 MpH = -log [H + ] = -log(2.839 x 10 -4 ) = 3.55b) What is the pH after the addition of 0.0050 moles of HCl?(Assume that the volume remains constant at 1.00 L.)• this is equivalent to adding 0.0050 M HCl to the buffer• adding a strong acid will increase the amount of theconjugate acid (e.g. formic acid) and decrease theamount of the conjugate base (e.g. formate ion)CHO 2 - + H + 6 HCHO 2[HCHO 2 ] = 0.550 M + 0.0050 M = 0.555 M[CHO 2 - ] = 0.350 M - 0.0050 M = 0.345 M[H + ] = K a X [HCHO 2 ]/[CHO 2 - ][H + ] = 1.8 x 10 -4 X (0.555)/(0.345) = 2.896 x 10 -4 MpH = -log [H + ] = -log(2.896 x 10 -4 ) = 3.54Note: slightly more acidic

c) What would be the pH after the addition of 0.0050 mol ofNaOH to the original buffer?• this is equivalent to adding 0.0050 M NaOH to the buffer• adding a strong base will increase the amount of theconjugate base (e.g. formate ion) and decrease theamount of the conjugate acid (e.g. formic acid)HCHO 2 + OH - 6 CHO 2 - + H 2 O[CHO 2 - ] = 0.350 M + 0.0050 M = 0.355 M[HCHO 2 ] = 0.550 M - 0.0050 M = 0.545 M[H + ] = K a X [HCHO 2 ]/[CHO 2 - ][H + ] = 1.8 x 10 -4 X (0.545)/(0.355) = 2.763 x 10 -4 MpH = -log [H + ] = -log(2.763 x 10 -4 ) = 3.56Note: slightly more basic