1. Calculating pH Chapter 16 part IV

2. Summary of General Strategies Think Chemistry: focus on solution and the components. It is usually easy to identify one reaction that is not important. Think Chemistry: focus on solution and the components. It is usually easy to identify one reaction that is not important. Be systematic: Acid-base problems require step by step approach. Be systematic: Acid-base problems require step by step approach. Be Flexible: Although all acid-base problems are similar, important differences do occur. Do not force a given problem into matching a problem you have solved before. Be Flexible: Although all acid-base problems are similar, important differences do occur. Do not force a given problem into matching a problem you have solved before.

3. Summary of General Strategies Be patient: the complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into workable steps. Be patient: the complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into workable steps. Understand & Think: don’t just memorize. Understand & Think: don’t just memorize.

4. Calculating the pH of a strong acid We need to look at the solutions components and their chemistry. We need to look at the solutions components and their chemistry. For example: 1.0 M HCL has virtually no HCl molecules as HCl ionizes “completely” into H+ and Cl-. For example: 1.0 M HCL has virtually no HCl molecules as HCl ionizes “completely” into H+ and Cl-. We need to determine which components are significant, and which can be ignored. We need to determine which components are significant, and which can be ignored.

5. Calculating the pH of a strong acid Find the major species: the solution components present in large amounts. Find the major species: the solution components present in large amounts. Back to our example: 1 M HCl Back to our example: 1 M HCl If it is 1 M than it must be a solute in a solvent to make a solution. Solvent is water. If it is 1 M than it must be a solute in a solvent to make a solution. Solvent is water. Major species are H+, Cl- & H 2 O. HCl is virtually dissociated and therefore not a major species. Major species are H+, Cl- & H 2 O. HCl is virtually dissociated and therefore not a major species.

6. Calculating the pH of a strong acid H+ is a major species from HCl, but what about H+ from H 2 O? H+ is a major species from HCl, but what about H+ from H 2 O? H 2 O  H + + OH - H 2 O  H + + OH - Is this an important source of H + ? Is this an important source of H + ? In pure water at 25°C [H + ]= 1X10 -7 M In pure water at 25°C [H + ]= 1X10 -7 M In water with 1.0 M HCl, water will produce even less H+. Why? In water with 1.0 M HCl, water will produce even less H+. Why?

7. Le Châtelier’s Principle Le Châtelier’s states that equilibrium will shift to reduce stress. If  [H+] is produced by HCl, then H 2 O produce less H+ is response to that stress. Le Châtelier’s states that equilibrium will shift to reduce stress. If  [H+] is produced by HCl, then H 2 O produce less H+ is response to that stress. Therefore: if 1 M [H+] from HCl, then is 1X10 -7 M[H+] from water significant? Therefore: if 1 M [H+] from HCl, then is 1X10 -7 M[H+] from water significant? No, pH=-log[1.0] No, pH=-log[1.0] pH =0.0 pH =0.0

8. Calculating the pH of a strong acid Calculate the pH of 0.1 M HNO 3. Calculate the pH of 0.1 M HNO 3. Major species: Major species: H + NO 3 - H 2 O HNO 3. H + NO 3 - H 2 O HNO 3. 0.1M0.1M 1X10 -7 M 0 0.1M0.1M 1X10 -7 M 0 pH = -log[H+] pH = -log[H+] pH = -log[0.1] pH = -log[0.1] pH=1.00 pH=1.00

9. Calculating the pH of a strong acid Calculate the pH of 1.0x10 -10 M HCl Calculate the pH of 1.0x10 -10 M HCl Major species: Major species: H + Cl - H 2 O HCl H + Cl - H 2 O HCl 1.0x10 -10 M 1.0x10 -10 M 1.0x10 -7 M 0 1.0x10 -10 M 1.0x10 -10 M 1.0x10 -7 M 0 pH=-log[H+] pH=-log[H+] What’s the sum of 1.0x10 -10 + 1.0x10 -7 What’s the sum of 1.0x10 -10 + 1.0x10 -7 pH=-log[1.0x10 -7 ] pH=-log[1.0x10 -7 ] pH=7.00 pH=7.00

10. Calculating the pH of a Weak acid Find the pH of a 1.00 M solution of HF, Ka= 7.2 X 10 -4. Find the pH of a 1.00 M solution of HF, Ka= 7.2 X 10 -4. Remember: Weak acids do not completely dissociate in water. Remember: Weak acids do not completely dissociate in water. Calculations with weak acids must be done carefully and systematically and are the prototype to almost all equilibrium problems. Following all the steps here make it easier to remember all the steps for more difficult problems. Calculations with weak acids must be done carefully and systematically and are the prototype to almost all equilibrium problems. Following all the steps here make it easier to remember all the steps for more difficult problems.

11. Calculating the pH of a Weak acid 1. Write the major species, Remember very little of the HF dissociates. 1. Write the major species, Remember very little of the HF dissociates. HF H 2 O HF H 2 O But which ones furnish the H+? But which ones furnish the H+? Answer: Both Answer: Both HF  H + + F-Ka = 7.2 X 10 -4 HF  H + + F-Ka = 7.2 X 10 -4 H 2 O  H + + OH-Kw = 1.0 X 10 -14 H 2 O  H + + OH-Kw = 1.0 X 10 -14

12. Calculating the pH of a Weak acid Which is the dominant source of H+? Which is the dominant source of H+? Ka = 7.2 X 10 -4 >> 1.0 X10 -14 Ka = 7.2 X 10 -4 >> 1.0 X10 -14 Thus the dominate source is HF. Thus the dominate source is HF. The equilibrium expression is: The equilibrium expression is: Ka = 7.2 X 10-4 = [H+][F-] Ka = 7.2 X 10-4 = [H+][F-] [HF] [HF]

13. Calculating the pH of a Weak acid Write the equilibrium expression. Write the equilibrium expression. Make your ice table Make your ice table I C E What are the [equilibrium]’s? What are the [equilibrium]’s? Does it meet the 5% rule? Does it meet the 5% rule? Are we done? Are we done?

14. Calculating the pH of a Weak acid The question asked for the pH!! The question asked for the pH!! [H+] was 2.7 X10 -2 [H+] was 2.7 X10 -2 pH=-log(2.7 X10 -2 ) pH=-log(2.7 X10 -2 ) pH= 1.57 pH= 1.57

15. Summary List all major species List all major species Chose species that produces H+ and write balanced equation. Chose species that produces H+ and write balanced equation. Using Ka +Kw for the reactions find the dominant source of H+ Using Ka +Kw for the reactions find the dominant source of H+ Write the equilibrium expression for the dominant source of H+ Write the equilibrium expression for the dominant source of H+ ICE in terms of X ICE in terms of X

16. Summary continued. Substitute the equilibrium [ ]’s into the equilibrium expression. Substitute the equilibrium [ ]’s into the equilibrium expression. Solve for X using the “easy” way when the Ka has a small value. HA-x=HA Solve for X using the “easy” way when the Ka has a small value. HA-x=HA Verify with the 5% rule Verify with the 5% rule Calculate [H+], pH Calculate [H+], pH

17. Example OCl- is a strong oxidizer in bleach and forms in chlorinated pools. OCl- is a strong oxidizer in bleach and forms in chlorinated pools. This relatively strong base forms a weak acid with a Ka = 3.5 X 10 -8 This relatively strong base forms a weak acid with a Ka = 3.5 X 10 -8 Find the pH of 0.100 M HOCl Find the pH of 0.100 M HOCl

18. Species, equations, ICE, etc

19. Answer pH=4.23 pH=4.23 What about a mixture of weak acids? What about a mixture of weak acids? Find the pH of a mixture of Find the pH of a mixture of 1.0 M HCN Ka= 6.2 X 10 -10 1.0 M HCN Ka= 6.2 X 10 -10 5.0 M HNO 2 Ka= 4.0 X 10 -4 5.0 M HNO 2 Ka= 4.0 X 10 -4 Also find the [CN-] at equilibrium Also find the [CN-] at equilibrium

20. Major Species, Equations, Ka expression, ICE

21. Answer: HCN, HNO 2, H 2 O HCN  H + + CN-Ka = 6.2 X 10 -10 HCN  H + + CN-Ka = 6.2 X 10 -10 H 2 O  H + + OH-Ka = 1.0 X 10 -14 H 2 O  H + + OH-Ka = 1.0 X 10 -14 HNO 2  H + + NO 2 -Ka = 1.0 X 10 -4 HNO 2  H + + NO 2 -Ka = 1.0 X 10 -4 Major [H+] is HNO2 Major [H+] is HNO2 I5.0 00 I5.0 00 C-X+X +X C-X+X +X E 5.0-X XX E 5.0-X XX Ka=4.0X10-4 =[H+][NO 2 -] ≈ X 2 /5 Ka=4.0X10-4 =[H+][NO 2 -] ≈ X 2 /5 [HNO 2 ] [HNO 2 ]

22. Answer continued X= 4.5 X 10-2 X= 4.5 X 10-2 5% rule (4.5 x10-2/5)X100 = 0.9% 5% rule (4.5 x10-2/5)X100 = 0.9% Valid Valid [H+] = 0.045 pH = 1.35 [H+] = 0.045 pH = 1.35 Find [CN-] Find [CN-]

23. HCN  H + + CN- Ka = 6.2 X 10 -10 This is the only source of CN- This is the only source of CN- Ka = 6.2 X 10 -10 =([H+][CN-])/[HCN] Ka = 6.2 X 10 -10 =([H+][CN-])/[HCN] HCN  H + + CN- HCN  H + + CN- I 1.0.045 0 I 1.0.045 0 C -X +X +X C -X +X +X E 1.0-X 0.045 + X X E 1.0-X 0.045 + X X 1.0 –X ≈ 1.0 1.0 –X ≈ 1.0 0.045 + X ≈0.045 0.045 + X ≈0.045

24. And Finally 6.2 x 10 -10 = 0.045 [CN-] 6.2 x 10 -10 = 0.045 [CN-] 1.0 1.0 [CN-]=6.2 X 10 -10 [CN-]=6.2 X 10 -10 0.045 0.045 [CN-]=1.4 X10 -8 M [CN-]=1.4 X10 -8 M