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Calculating pH

Calculating pH. Chapter 16 part IV. Summary of General Strategies. 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.

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Calculating pH

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  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. • 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.

  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. • 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. • 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.

  5. Calculating the pH of a strong acid • Find the major species: the solution components present in large amounts. • 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. • Major species are H+, Cl- & H2O. 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 H2O? • H2O  H+ + OH- • Is this an important source of H+? • In pure water at 25°C [H+]= 1X10-7M • 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 H2O produce less H+ is response to that stress. • Therefore: if 1 M [H+] from HCl, then is 1X10-7M[H+] from water significant? • No, pH=-log[1.0] • pH =0.0

  8. Calculating the pH of a strong acid • Calculate the pH of 0.1 M HNO3. • Major species: • H+ NO3- H2O HNO3. • 0.1 M 0.1M 1X10-7M 0 • pH = -log[H+] • pH = -log[0.1] • pH=1.00

  9. Calculating the pH of a strong acid • Calculate the pH of 1.0x10-10M HCl • Major species: • H+ Cl- H2O HCl • 1.0x10-10M 1.0x10-10M 1.0x10-7M 0 • pH=-log[H+] • What’s the sum of 1.0x10-10+ 1.0x10-7 • pH=-log[1.0x10-7] • 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. • 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.

  11. Calculating the pH of a Weak acid • 1. Write the major species, Remember very little of the HF dissociates. • HF H2O • But which ones furnish the H+? • Answer: Both • HFH+ + F- Ka = 7.2 X 10-4 • H2O  H+ + OH- Kw = 1.0 X 10-14

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

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

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

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

  16. Summary continued. • 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 • Verify with the 5% rule • Calculate [H+], pH

  17. Example • 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 • Find the pH of 0.100 M HOCl

  18. Species, equations, ICE, etc

  19. Answer • pH=4.23 • What about a mixture of weak acids? • Find the pH of a mixture of • 1.0 M HCN Ka= 6.2 X 10-10 • 5.0 M HNO2 Ka= 4.0 X 10-4 • Also find the [CN-] at equilibrium

  20. Major Species, Equations, Ka expression, ICE

  21. Answer: HCN, HNO2, H2O • HCN  H+ + CN- Ka = 6.2 X 10-10 • H2O  H+ + OH- Ka = 1.0 X 10-14 • HNO2 H++ NO2- Ka = 1.0 X 10-4 • Major [H+] is HNO2 • I 5.0 0 0 • C -X +X +X • E 5.0-X X X • Ka=4.0X10-4 =[H+][NO2-] ≈ X2/5 • [HNO2]

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

  23. HCN  H+ + CN- Ka = 6.2 X 10-10 • This is the only source of CN- • Ka = 6.2 X 10-10 =([H+][CN-])/[HCN] • HCN  H+ + CN- • I 1.0 .045 0 • C -X +X +X • E 1.0-X 0.045 + X X • 1.0 –X ≈ 1.0 • 0.045 + X ≈0.045

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

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