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Acids Lesson 12 Calculating Ka From pH
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid.
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H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C E
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C E pH=1.28
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C E pH=1.28 [H + ]=10 -1.28
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C E pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C E0.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C - 0.052480.052480.05248 E0.047520.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C - 0.052480.052480.05248 E0.047520.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M Ka=[H + ][HC 2 O 4 - ] [H 2 C 2 O 4 ]
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C - 0.052480.052480.05248 E0.047520.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M Ka=[H + ][HC 2 O 4 - ] =(0.05248) 2 [H 2 C 2 O 4 ] 0.04752
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1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C - 0.052480.052480.05248 E0.047520.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M Ka=[H + ][HC 2 O 4 - ] =(0.05248) 2 = 5.8 x 10 -2 [H 2 C 2 O 4 ] 0.04752
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb.
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NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E pH=11.427
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E pH=11.427 pOH=2.573
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E pH=11.427 pOH=2.573 [OH - ]=10 -2.573
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E0.0026730.002673 pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [H + ]=0.002673 M
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C - 0.0026730.0026730.002673 E0.39730.0026730.002673 pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C - 0.0026730.0026730.002673 E0.39730.0026730.002673 pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M Kb=[NH 4 + ][OH - ] [NH 3 ]
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C - 0.0026730.0026730.002673 E0.39730.0026730.002673 pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M Kb=[NH 4 + ][OH - ] =(0.002673) 2 [NH 3 ] 0.3973
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2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C - 0.0026730.0026730.002673 E0.39730.0026730.002673 pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M Kb=[NH 4 + ][OH - ] =(0.002673) 2 = 1.8 x 10 -5 [NH 3 ] 0.3973
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3.The pH of a 1.0 M triprotic weak acid is 4.568. Calculate the Ka and identify the acid. Ka=7.3 x 10 -10 Boric acid
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -.
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CN - + H 2 O ⇄ HCN+OH -
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000 C0.002858 0.002858 0.002858
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000 C0.002858 0.002858 0.002858 E0.1971 0.002858 0.002858
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000 C0.002858 0.002858 0.002858 E0.1971 0.002858 0.002858 [OH - ] =0.002858 M
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000 C0.002858 0.002858 0.002858 E0.1971 0.002858 0.002858 [OH - ] =0.002858 M Kb=[HCN][OH - ] = [CN - ]
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4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000 C0.002858 0.002858 0.002858 E0.1971 0.002858 0.002858 [OH - ] =0.002858 M Kb=[HCN][OH - ] =(0.002858) 2 = 4.1 x 10 -5 [CN - ] 0.1971
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5.Calculate the pH of 0.020 M H 3 BO 3 H 3 BO 3 ⇄ H + +H 2 BO 3 - I0.020 M00 CxxxCxxx E0.020 - xxx 0 small ka x 2 =3.8 x 10 -10 0.020 x=[H + ]=2.76 x 10 -6 M pH=-Log[2.76 x 10 -6 ] pH=5.42 2 sig figs due to molarity and Ka
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6.Calculate the pH of a solution made by mixing 100.0 mL of 0.050 M HCl with 100 mL of water. HCl→H + +Cl - 1(0.050 M)0.025 M0.025 M 2 pH=1.60
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