Lec 15 Feb 06 Fri - = x =[H 3 O pHinit = –log1.8x10 –5...

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2.13 (1) How will pH change when 0.020 mol of HCl is added to water to make 1 L of solution? HCl is a strong acid Review and Introduction to Buffers HCl + H 2 O H 3 O + + Cl - (completely dissociated) [H3O+] = [HCl] = (0.020 mol)/(1 L) = 0.020 mol/L = 0.020 M pH = - log (0.020) = 1.70 Initial pH = 7 (pure water at 25 deg C) pH = pHinit – pHfinal = 7.00 – 1.70 = 5.30 Change in pH CH 3 COOH (aq) + H 2 O (l) CH 3 COO (aq) + H 3 O + (aq) I 0.50 --- 0.50 0 C -x --- +x +x E 0.50-x --- 0.50+x x (2) How will pH change when 0.020 mol of HCl is added to 1 L of solution that is 0.590 M CH3COOH and 0.50 M CH3COONa? Initial pH = pH of a mixture of weak acid and its conjugate base K a = 1.8 ! 10 " 5 = [H 3 O + ][CH 3 COO " ] [CH 3 COOH] = ( x )(0.50 + x ) 0.50 " x Using approximation x << 0.50: ( x ) 0.50 ! 100% = 1.8 ! 10 " 5 0.50 ! 100% = 3.6 ! 10 " 3 < 5% K a = 1.8 ! 10 " 5 = ( x )(0.50) 0.50 =
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Unformatted text preview: = x = [H 3 O + ] pHinit = –log1.8x10 –5 = 4.74 2.14 CH 3 COO − (aq) + H 3 O + (aq) ⇔ CH 3 COOH (aq) + H 2 O (l) I 0.50 0.020 0.50---C-0.020 -0.020 +0.020---F 0.048 0.0 0.520---pH when HCl is added to a mixture of weak acid and its conjugate base K a = 1.8 ! 10 " 5 = ( x )(0.48 + x ) 0.52 " x # ( x )(0.48) 0.52 Notice that frst H3O+ will react with base to Form acid: CH 3 COOH (aq) + H 2 O (l) ⇔ CH 3 COO − (aq) + H 3 O + (aq) I 0.52---0.48 C-x---+x +x E 0.52-x--- 0.48+x x Now we have acid/base equilibrium (with new initial concentrations) x = 2.0x10 –5 M pH = –log 2.0x10 –5 = 4.70 pH = pHinit – pHfinal = 4.74 – 4.70 = 0.04 Change in pH We have a BUFFER!...
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