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CH 201 Self Study Worksheet 7A - Key 1. a.What is the pH of a solution that is 0.25 M in HF and 0.15 M F1-? The pKa of HF is 3.14. pH = 3.14 + log(0.15 M)/(0.25 M) = 2.92 [base] < [acid] so the pH should be less than the pKa. b. Write out the chemical reaction for when NaOH is added to this HF/F1- solution? HF + OH1-  F1- + H2O The acid neutralizes all of the OH1- that is added. c. Upon addition of NaOH to this buffer solution, would the pH increase or decrease? Increase, but only a little! (Think about what your answers should be.) d. Calculate the pH of the solution after addition of 10.00 mL of 0.050M NaOH to 90.00 mL of this buffer solution. HF + OH1-  F1- + H2O Reacts completely! 22.5 mmol 0.5 mmol 13.5 mmol -0.5 -0.5 +0.5 22.0 mmol 0 mmol 14.0 mmol 0.22 M 0.14 M pH = 3.14 + log(14.0 mmol)/(22.0) = 2.94 OR pH = 3.14 + log(0.14 M)/(0.24 M) = 2.94 e. Over what pH range is HF/F1- a good buffer? 2.14 – 4.14, pKa 1 2. Phosphate buffers are often used in biochemistry labs. The most common buffer system involves H2PO41- and HPO42-. H2PO41- has a pKa of 7.21. The purpose of this question is to create a buffer of pH 7.00 using three possible methods. a. Direct Method: How many grams of NaH2PO4 must be added to 500. mL of a 0.400 M HPO42- to produce a buffer of pH 7.00? Mm NaH2PO4 is 119.98 g/mol. 7.00 = 7.21 + log(500 ml x 0.400 mmol HPO42-/ mmol H2PO41-) 10-0.21 = 200 mmol HPO42-/ mmol H2PO41- mmol H2PO41- = 324.4 mmol 324.4 mmol NaH2PO4 x (1 mol/1000 mmol ) x (119.98 g/mol) = 38.9 g b. Addition of strong base to weak acid: How many ml of 3.00 M NaOH must be added to 500. mL of a 0.400 M H2PO41- solution to produce a buffer of pH 7.00? H2PO41- + OH1-  HPO42- + H2O Reacts completely! 200. mmol x 0 -x -x +x 200-x mmol 0 mmol x mmol 7.00 = 7.21 + log(x mmol HPO42-)/(200-x H2PO41-) 10-0.21 = x mmol HPO42-/ 200 – x mmol H2PO41- 123.32 – 0.6166x = x 123.32 = 1.6166x  x = 76.28 mmol OH1- 76.28 mmol OH1- x ( ml / 3.00 mmol OH1-) = 25.4 mL OH1-CH 201 Self Study Worksheet 7A - Key c. Addition of strong acid to weak base: How many ml of 3.00 M HCl must be added to 500. mL of a 0.400 M HPO42- solution to produce a buffer of pH 7.00? HPO42- + H3O1+  H2PO41- + H2O Reacts completely! 200. mmol x 0 -x -x +x 200-x mmol 0 mmol x mmol 7.00 = 7.21 + log(200-x mmol HPO42-)/( x H2PO41-) 10-0.21 = 200 - x mmol HPO42-/ x mmol H2PO41- 0.6166x = 200 - x 1.6166x = 200  x = 123.7 mmol H3O1+ 123.7 mmol H3O1+ x ( ml / 3.00 mmol H3O1+) = 41.2 mL H3O1+ 3. Consider the titration of 25.00 mL of 0.0800 M NaOH with 0.1000 M HCl. a. What is the equivalence volume for this titration? (25.00 mL OH1-)(0.0800 M OH1-)(1 mmol H3O1+/1 mmol OH1-)/(0.1000 M H3O1+) = 20.00 mL HCl b. What is the pH of the 0.0800 M base solution before any titrant is added? NaOH is a strong base so all of it dissociates to OH1-. [OH1-] = 0.0800 M  pOH = 1.10  pH = 12.90 c. What is the pH after 4.00 mL of titrant have been added? (4.00 mL)(0.1000 M) = 0.400 mmol H3O1+ added (25.00 mL)(0.0800 M NaOH) = 2.00 mmol OH1- originally OH1- + H3O1-  2H2O initial: 2.00 mmol 0.40 mmol ∆: -0.4 mmol -0.4 mmol final: 1.6 mmol 0 mmol Strong base in excess: [OH1-] = (1.6 mmol)/(25.00 mL + 4.00 mL) = 0.0552 M pOH = 1.26  pH = 12.74 d. What is the pH at the equivalence point? (20.00 mL)(0.1000 M) = 2.00 mmol H3O1+ added OH1- + H3O1-  2H2O initial: 2.00 mmol 2.00 mmol ∆: -2.00 mmol-2.00 mmol final: 0 mmol 0 mmol No strong acid or base in excess, pure water!  pH = 7.00 e. What is the pH after 22.00 mL of titrant have been added? (22.00 mL)(0.1000 M) = 2.20 mmol H3O1+ added OH1- + H3O1-  2H2O initial: 2.00 mmol 2.20 mmol ∆: -2.00 mmol-2.00 mmol final: 0.0 0 mmol 0.20 mmol [H3O1+] = (0.200 mmol)/(25 mL + 22 mL) = 4.26x10-3 pH = 2.37


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