1. Lecture 183/4/05 Spring Break QUIZ

2. Quiz 6 1. What is the pH of a buffer solution containing 0.3 M HNO 2 and 0.25 M NaNO 2 ? K a (HNO 2 ) = 4.5 x 10 -4 2.Place the following in order of high pH to low pH. NaCl, NaH, Na 2 CO 3

3. Titration Curve Why do you use one? Equivalence point vs. Endpoint pH vs. acid/base added

6. Building a titration curve via calculations 4 Regions (to build one from calculations): 1) initial pH 2) before equivalence point 3) at equivalence point 4) after equivalence point

7. Building a titration curve via calculations Comparison of a titration of a strong acid vs. a weak acid Regionx-axis (mL of acid/base) 0.1 M NaOH y-axis (pH) 100 mL of 0.1 M HCl y-axis (pH) 100 mL of 0.1 M HF Initial pH Before the equivalence point (1/2 equiv. point) equivalence point After the equivalence point

8. Initial pH (NaOH titrating HCl) X-axis = 0 mL (No base has been added) Y-axis: solve for pH 0.1 M HCl  0.1 M H + pH = -log(0.1) = 1

9. Building a titration curve via calculations Comparison of a titration of a strong acid vs. a weak acid Regionx-axis (mL of acid/base) 0.1 M NaOH y-axis (pH) 100 mL of 0.1 M HCl y-axis (pH) 100 mL of 0.1 M HF Initial pH 01 Before the equivalence point (1/2 equiv. point) equivalence point After the equivalence point

11. Equivalence point (NaOH titrating HCl) X-axis moles OH - added = moles H 3 O + in solution C OH V OH = C H V H (0.1 M)V OH = (0.1 M)(100 mL) V OH = 100 mL Y-axis Neutralization of strong acid by strong base pH = 7

12. Building a titration curve via calculations Comparison of a titration of a strong acid vs. a weak acid Regionx-axis (mL of acid/base) 0.1 M NaOH y-axis (pH) 100 mL of 0.1 M HCl y-axis (pH) 100 mL of 0.1 M HF Initial pH 01 Before the equivalence point (1/2 equiv. point) equivalence point 1007 After the equivalence point

14. ½ Equivalence point (NaOH titrating HCl) X-axis moles OH - added = ½ moles H 3 O + in solution V equivalence = 100 mL, so V 1/2equivalence = 50 mL Y-axis H + + OH -  H 2 O (0.1 L)(0.1 M) (0.05 L)(0.1 M) 0.01 moles0.005 moles 0.005 moles0 moles Total volume = volume of acid solution + volume of base added = 150 mL pH = -log (0.005 moles/0.15 L) = 1.47

15. Building a titration curve via calculations Comparison of a titration of a strong acid vs. a weak acid Regionx-axis (mL of acid/base) 0.1 M NaOH y-axis (pH) 100 mL of 0.1 M HCl y-axis (pH) 100 mL of 0.1 M HF Initial pH 01 Before the equivalence point (1/2 equiv. point) 501.47 equivalence point 1007 After the equivalence point

17. After Equivalence point (NaOH titrating HCl) X-axis Can pick any volume greater than V equivalence Often pick 3/2 equivalence volume = 150 mL Y-axis H + + OH -  H 2 O (0.1 L)(0.1 M) (0.15 L)(0.1 M) 0.01 moles0.015 moles 0 moles0.005 moles Total volume = volume of acid solution + volume of base added = 250 mL pOH = -log (0.005 moles/0.25 L) = 1.7 pH = 14 – 1.7 = 12.3

18. Building a titration curve via calculations Comparison of a titration of a strong acid vs. a weak acid Regionx-axis (mL of acid/base) 0.1 M NaOH y-axis (pH) 100 mL of 0.1 M HCl y-axis (pH) 100 mL of 0.1 M HF Initial pH 01 Before the equivalence point (1/2 equiv. point) 501.47 equivalence point 1007 After the equivalence point 15012.3