1. Ka, Kb

2. Comparing the pH of two acids 1.Predict the pH of HCl and HF (below) 2.Calibrate a pH meter 3.Measure the pH of HCl(aq) and HF(aq) 4.Complete the chart below HCl (aq)HF (aq) [ ] in mol/L (on label) Net ionic equation Predicted [H + ] Predicted pH pH measured Actual [H + ] Conductivity (demo)Higher / strongerLower / weaker 0.05 HCl H + + Cl – HF H + + F – 0.05 -log(0.05)=1.3 1.32.3 ? 10 –pH = 0.05 10 –pH = 0.005

3. Questions Read15.3. (pg. 607+) 1.Based on your results, which acid ionizes (forms ions) to a greater degree? 2.Which two measurements taken in the lab support your answer to 1? 3.What is another name for Ka? 4.Solve PE 5, 6 5.Write the Ka equation for HCl (aq) and HF (aq) from todays lab 6.Solve for PE 8, 9 (use this equilibrium for butyric acid: HBu H + + Bu – ) 7.For HF(aq) set up a RICE chart, then solve for Ka. How does your value for Ka compare to the accepted value (pg. 608)? 8.Try PE 10 (follow example 15.7 on pg. 610)

4. Answers 1.HCl ionizes more than HF 2.HCl has a lower pH (indicating more H + ), & a higher conductivity (indicating more ions) 3.Ka: acid ionization constant 4.HNO 2 H + + NO 2 –, Ka=[H + ][NO 2 – ]/[HNO 2 ] HPO 4 2– H + + PO 4 3–,Ka=[H + ][PO 4 3– ]/[HPO 4 2– ] 5.HCl H + + Cl –, Ka=[H + ][Cl – ]/[HCl] HF H + + F –, Ka=[H + ][F – ]/[HF]

5. PE 8 - pg. 610 HBu H + + Bu – R I C E HBuH+H+ Bu – 111 0.010000 -0.0004+0.0004 0.00960.0004 [HBu] Ka = [H + ][Bu – ] = [.0096] [0.0004] 2 = 1.67 x 10 – 5 [H + ] = 10 – pH = 10 – 3.40 = 3.98 x 10 – 4

6. PE 9 - pg. 610 HBu H + + Bu – R I C E HBuH+H+ Bu – 111 0.010000 -0.001+0.001 0.0090.001 [HBu] Ka = [H + ][Bu – ] = [.009] [0.001] 2 = 1.1 x 10 – 4 [H + ] = 10 – pH = 10 – 2.98 = 1.05 x 10 – 3

7. Question 7: HF H + + F – R I C E HFH+H+ F–F– 111 0.0500 -0.005+0.005 0.0450.005 [HF] Ka = [H + ][F – ] = [.045] [0.005] 2 = 5.6 x 10 – 4 [H + ] = 10 – pH = 10 – 2.3 = 0.005 Accepted value of Ka for HF is 6.4 x 10 – 4

8. 10: HC 2 H 4 NO 2 H + + C 2 H 4 NO 2 – R I C E HC 2 H 4 NO 2 H+H+ C 2 H 4 NO 2 – 111 0.01000 -x+x 0.010 - xxx [HF] Ka = [H + ][C 2 H 4 NO 2 – ] = [0.010 - x] [x] 2 = 1.4 x 10 – 5 Since x is small 0.010 – x = 0.010 x= 3.74 x 10 –5 M, pH = 3.43 [0.010] [x] 2 =1.4 x 10 – 5

9. Ka summary Ka follows the pattern of other K equations I.e. for HA(aq) + H 2 O(l) H 3 O + (aq) + A – (aq) Ka = [H 3 O + ][A – ] / [HA] Notice that H 2 O is ignored because it is liquid HA cannot be ignored because it is aqueous This is different than with Ksp. In Ksp, solids could only be in solution as ions Acids can be in solution whether ionized or not The solubility of acids makes sense if you think back to the partial charges in HCl for ex.

10. Ka summary Generally Ka tells you about acid strength Strong acids have high Ka values A strong acid is an acid that completely ionizes. E.g. HCl + H 2 O H 3 O + + Cl – A weak acid is an acid that doesnt ionize completely. E.g. HF + H 2 O H 3 O + + F – Note: dont get confused between strength and concentration. 1 M HCN has a smaller [H + ], thus a higher pH, than 0.001 M HCl In general:Ka < 10 – 3 Weak acid 10 – 3 < Ka < 1Moderate acid Ka > 1 Strong acid

11. Dissociation vs. Ionization Ionization and dissociation indicate ions form Dissociation: ions form when a chemical comes apart. E.g. NaCl melts to form Na +, Cl – Ionization: ions form when two chemicals react. E.g. HCl(aq) + H 2 O H 3 O + (aq) + Cl – (aq) Even though we write HCl H + + Cl –, this is just an abbreviation. In reality HCl reacts with H 2 O, thus it is an ionization not a dissociation Note that NaCl can also dissociate in water. This is not an ionization, since water is only required to stabilize ions (it is not needed as a reactant involved in forming ions)

12. Kb – the last K (I promise) Kb is similar to Ka except b stands for base The general reaction involving a base can be written as B(aq) + H 2 O BH + (aq) + OH – (aq) Thus Kb = [BH + ] [OH – ] / [B] Recall: shorthand for Ka is HA H + + A – Kb has no shorthand form Read pg. 614 - 617 Try PE 12 (a-c), 13, 14 (for 13, you do not need to know the chemical formula of morphine. Symbolize it with M)

13. PE 12 a)CN – (aq) + H 2 O HCN(aq) + OH – (aq) K b = [HCN][OH – ] / [CN – ] b)C 2 H 3 O 2 – (aq) + H 2 O HC 2 H 3 O 2 (aq) + OH – (aq) K b = [HC 2 H 3 O 2 ][OH – ] / [C 2 H 3 O 2 – ] c)C 6 H 5 NH 2 (aq) + H 2 O C 6 H 5 NH 3 + (aq) + OH – (aq) K b = [C 6 H 5 NH 3 + ][OH – ] / [C 6 H 5 NH 2 ]

14. PE 13 - pg. 617 M + H 2 O MH + + OH – R I C E MMH + OH – 111 0.01000 -0.00013+0.00013 0.00013 0.00987 [M] Kb = [MH + ] [OH – ] = =1.7 x 10 -6 pOH = 14 - pH = 14 - 10.10 = 3.90 [OH-] = 10 -pOH = 10 -3.90 = 1.26 x 10 -4 [0.00987] [0.00013]

15. PE 14 - pg. 617 M + H 2 O MH + + OH – R I C E NH 3 NH 4 + OH – 111 0.02000 -x+x xx0.020 - x [0.020] Kb = [x] = = 1.8 x 10 -5 x= 6.0 x 10 -4 pOH = -log[OH - ] = 3.22 pH = 14 - pOH = 10.78 [0.020] x2x2

16. Strength of conjugates Consider HCl(l) + H 2 O Cl – (aq) + H 3 O + (aq) The Ka for HCl is [Cl – (aq)] [H 3 O + (aq)] / [HCl(aq)] Also, Cl – (aq) + H 2 O(aq) HCl(l) + OH – The Kb for Cl – is [HCl(aq)] / [Cl – (aq)] [H 3 O + (aq)]

17. Relative values of Ka Recall for HX H + + X –, Ka = [H+][X – ] / [HX] Q - what does a large Ka indicate? A - equilibrium is far to the right (all dissociates) Thus a large Ka = strong acid Look at Table 15.4 on page 608 The text uses this definition: Ka < 10 –3 is a weak acid 10 –3 < Ka < 1 is a moderate acid 1 < Ka is a strong acid These definitions are somewhat arbitrary, we will not focus on this. Just remember a high Ka means the acid is strong. For more lessons, visit www.chalkbored.com www.chalkbored.com