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Entry Task: February 14 th Monday

Entry Task: February 14 th Monday. Sketch this on task sheet and what type of titration graph is it? . Agenda. Sign off and discuss Buffer & T-curves pH, Buffer and Alka Seltzer Lab Start on calculating pH in titration NOTES.

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Entry Task: February 14 th Monday

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  1. Entry Task: February 14th Monday Sketch this on task sheet and what type of titration graph is it?

  2. Agenda • Sign off and discuss Buffer & T-curves • pH, Buffer and Alka Seltzer Lab • Start on calculating pH in titration NOTES

  3. 1. What substance might you mix with HC2H3O2 to produce a buffered solution? Explain how the buffer would work with the addition of an acid or base. NaC2H3O2 or any salt with C2H3O2- If an addition of an acid (H+) will get tied up with the acetate ions not building up in the solution. Same idea with a base.

  4. 2. Using appropriate chemical equations, show how a mixture of H2PO4- and HPO42-, acting as conjugate acid and base respectively, can respond to the addition of the following.a) HClb) NaOH + HPO4-2 Cl- + H2PO4- + H2PO4- H2O + Na+ + HPO4-2

  5. 3. Could you create a buffer by mixing HI (aq) and NaI (aq)? Explain your answer No, HI is a strong acid and no equilibrium for a buffer solution

  6. 4. A buffer is prepared by adding 20.0 g of acetic acid, HC2H3O2, and 20.0 g of sodium acetate, NaC2H3O2 to enough water to form 2.0L of solution.a) Determine the pH of the buffer HC2H3O2  H + + C2H3O2- Calculate the molarity first then put it in an Ka expression to solve for pH. 20.0 g/60g in 1 mole = 0.33 moles of HC2H3O2 in 2.0 L = 0.167M 20.0 g/82g in 1 mole = 0.30 moles of NaC2H3O2in 2.0 L = 0.122M 1.8 x 10-5= [x][0.122] [0.167] Get x alone and solve 2.497 x10-5 = x= [H+] pH= 4.6

  7. 4. A buffer is prepared by adding 20.0 g of acetic acid, HC2H3O2, and 20.0 g of sodium acetate, NaC2H3O2 to enough water to form 2.0L of solution.b) Calculate the pH after 0.50 mol of NaOH is added to the 2.0L solution (assume no volume change) HC2H3O2  H + + C2H3O2-

  8. Calculating pH Changes in Buffers Calculate the pH after 0.50 mol of NaOH is added to the 2.0L solution (assume no volume change) HC2H3O2  OH - + C2H3O2- Added neutralizes any H+ ions

  9. Calculating pH Changes in Buffers Ka expression with NEW molarities to calculate new pH 1.8 x 10-5= [x][0.622] [0.333] Get x alone and solve 9.6 x10-6 = x= [H+] pH= 5.0

  10. 4. A buffer is prepared by adding 20.0 g of acetic acid, HC2H3O2, and 20.0 g of sodium acetate, NaC2H3O2 to enough water to form 2.0L of solution.c) Calculate the pH after 0.50 mol of HCl is added to the 2.0L solution (assume no volume change) HC2H3O2  H + + C2H3O2-

  11. Calculating pH Changes in Buffers Calculate the pH after 0.50 mol of HCl is added to the 2.0L solution (assume no volume change) HC2H3O2  H + + C2H3O2- • Added neutralizes any C2H3O2-ions

  12. Calculating pH Changes in Buffers Ka expression with NEW molarities to calculate new pH 1.8 x 10-5= [x][0.378] [0.667] Get x alone and solve 3.176 x10-5 = x= [H+] pH= 4.5

  13. 5. Calculate the pH of a solution formed by mixing 35 mL of 0.20M NaHCO3 with 55mL of 0.15M Na2CO3 if Ka2 of H2CO3 is 5.6 x 10-11. TOTAL VOLUME? 90 mls We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the same!! (0.20M)(0.035L)=(x) (0.90L)= 7.78x10-3 M of NaHCO3(new M) (0.15M)(0.055L)=(x)(0.90L)= 9.17x10-3MofNa2CO3 (new M)

  14. 5. Calculate the pH of a solution formed by mixing 35 mL of 0.20M NaHCO3 with 55mL of 0.15M Na2CO3 if Ka2 of H2CO3 is 5.6 x 10-11. Plug Back into Ka expression 5.6 x 10-11 = [x][9.17x10-3] [7.78x10-3] Rearrange to get X by itself x = [H+]= 4.75 x 10-11 pH = -log(6.32 x 10-11) = 10.3

  15. 6. Calculate the pH of a 500ml buffer solution that is 0.125M of HC2H3O2 and 0.115 M of NaC2H3O2. The Ka is 1.8 x10-5. 1.8 x 10-5= [x][0.115] [0.125] Get x alone and solve 1.96 x10-5 = x= [H+] pH= 4.7

  16. 6. Calculate the pH of a 500ml buffer solution that is 0.125M of HC2H3O2 and 0.115 M of NaC2H3O2. The Ka is 1.8 x10-5.Now calculate its pH after 0.01 M of HCl has been added 1.8 x 10-5= [x][0.105] [0.135] Get x alone and solve 2.3 x10-5 = x= [H+] pH= 4.64

  17. Titration Curves Information from the Curve: There are several things you can read from the titration curve itself. Consider this titration curve weak 1. This is a ___________ (strong/weak) acid titrated with a strong base. The acid is ________________ (monoprotic/diprotic). How would the other strength of acid look? mono If it was a strong acid the curve would be more “sharp”. • Place a dot () on the curve at the equivalence point. The pH at the equivalence point is ____. Choose a good indicator for this titration from Figure 16.7 on page 604 of your textbook. • 3. What volume of base was used to titrate the acid solution? _______ mL • 4. Place a box () on the curve where the pH of the solution = the pKa of the acid.What is the pH at this point? _______ • What is the pKa of the acid? _______ • What is the Ka of the acid? _____________ ~8.8 Phenolphthalein 25 mls ~4.8 ~4.8 1.58 x 10-5

  18. Titration Curves Information from the Curve: There are several things you can read from the titration curve itself. Consider this titration curve 1. Identify the solutions involved in this titration. A strong base in a weak acid • Place a dot () on the curve at the equivalence point. The pH at the equivalence point is ____. Choose a good indicator for this titration from Figure 16.7 on page 604 of your textbook. • 3. What volume of base was used to titrate the acid solution? _______ mL • 4. Place a box () on the curve where the pH of the solution = the pKbof the base.What is the pH at this point? _______ • What is the pKbof the acid? _______ • What is the Kbof the acid? _____________ ~5.8 Methyl red 20 mls ~8.8 ~8.8 1.58 x 10-9

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