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Acids Lesson 12 Calculating Ka From pH. 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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Presentation on theme: "Acids Lesson 12 Calculating Ka From pH. 1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid."— Presentation transcript:

1 Acids Lesson 12 Calculating Ka From pH

2 1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid.

3 H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 00 C E

4 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

5 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

6 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

7 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

8 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

9 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 ]

10 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

11 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

12 2.If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb.

13 NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 00 C E

14 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

15 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

16 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

17 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

18 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

19 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

20 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 ]

21 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

22 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

23 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

24 4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -.

25 CN - + H 2 O ⇄ HCN+OH -

26 4.The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.2000

27 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

28 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

29 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

30 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 - ]

31 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

32 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

33 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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