Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Published byDarcy Clark Modified over 8 years ago

Similar presentations

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

Download ppt "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."

Similar presentations

Salts and pH. Soluble salts dissociate in water to produce ions. Salts are basically ionic compounds that can be formed from the reaction from an acid.

Salts and pH. Soluble salts dissociate in water to produce ions. Salts are basically ionic compounds that can be formed from the reaction from an acid.
Chapter 14 Arrhenius –Acid – create H + in water –Base – create OH - in water Bronsted-Lowery –Acid – donates proton (H + ) –Base – accepts proton (H +

Chapter 14 Arrhenius –Acid – create H + in water –Base – create OH - in water Bronsted-Lowery –Acid – donates proton (H + ) –Base – accepts proton (H +
1 Acetic acid, has a K a of 1.7 x 10 -5. Determine the pH of a 0.10 M solution of acetic acid. Hint: First write out the equilibrium expression of the.

1 Acetic acid, has a K a of 1.7 x Determine the pH of a 0.10 M solution of acetic acid. Hint: First write out the equilibrium expression of the.
Acid-Base Reactions Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should.

Acid-Base Reactions Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part of the work should.
8.7 Acid-Base Titration Learning Goals … … determine the pH of the solution formed in a neutralization reaction.

8.7 Acid-Base Titration Learning Goals … … determine the pH of the solution formed in a neutralization reaction.
Relationship Between Ka and Kb. Consider the dissociation of a weak acid: CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO - (aq) K a = [H 3 O + ][CH.

Relationship Between Ka and Kb. Consider the dissociation of a weak acid: CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO - (aq) K a = [H 3 O + ][CH.
Lecture 122/12/07. pH What is it? How do you measure it?

Lecture 122/12/07. pH What is it? How do you measure it?
Acids & Bases Lesson 13 Weak Base/Strong Acid calculations.

Acids & Bases Lesson 13 Weak Base/Strong Acid calculations.
Acid/Base Indicators Substance that changes color in the presence of an acid or a base – Red or Blue Litmus – Phenolphthalein (phth) – Bromothymol blue.

Acid/Base Indicators Substance that changes color in the presence of an acid or a base – Red or Blue Litmus – Phenolphthalein (phth) – Bromothymol blue.
Calculating the pH of Acids and Bases Strong vs. Weak.

Calculating the pH of Acids and Bases Strong vs. Weak.
Neutralization & Titrations

Neutralization & Titrations
Acids & Bases Hydrolysis Lesson 7. Hydrolysis Hydrolysis of a salt is… The reaction between water and the cation or anion ( or both) that make up the.

Acids & Bases Hydrolysis Lesson 7. Hydrolysis Hydrolysis of a salt is… The reaction between water and the cation or anion ( or both) that make up the.
Review of Acids. HClStrong Acid HCl  H + + Cl -0.10 M pH = -Log[H + ] =1.0 No ICE HFWeak Acid HF ⇌ H + + F - I0.10 M00 CxxxCxxx E0.10 - xxx small Ka.

Review of Acids. HClStrong Acid HCl  H + + Cl M pH = -Log[H + ] =1.0 No ICE HFWeak Acid HF ⇌ H + + F - I0.10 M00 CxxxCxxx E xxx small Ka.
Strength of Acids and Bases Do they ionize 100%?.

Strength of Acids and Bases Do they ionize 100%?.
Entry Task: Jan 22 nd Thursday Turn in Determine Ka Lab.

Entry Task: Jan 22 nd Thursday Turn in Determine Ka Lab.
Weak acid and base calculations. What’s so hard? Unlike strong acids and bases, weak examples do not dissociate fully. For example, a 1 molL -1 HCl solution.

Weak acid and base calculations. What’s so hard? Unlike strong acids and bases, weak examples do not dissociate fully. For example, a 1 molL -1 HCl solution.
Acids Lesson 8 Application of Hydrolysis pH Calculations.

Acids Lesson 8 Application of Hydrolysis pH Calculations.
Acids Lesson 8 Weak Acids pH, Ka Calculations.

Acids Lesson 8 Weak Acids pH, Ka Calculations.
Acids & Bases Part 2. Acid & Base Ionization Constants A weak acid or base produces a reaction that only partially goes forward. The acid or base ionization.

Acids & Bases Part 2. Acid & Base Ionization Constants A weak acid or base produces a reaction that only partially goes forward. The acid or base ionization.
Acids & Bases Lesson 9 pH, pOH, Kb for Weak Bases.

Acids & Bases Lesson 9 pH, pOH, Kb for Weak Bases.

Similar presentations

Ads by Google