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- 13. A 60.00 mL sample of 0.075 M sodium benzoate (NaC7H5O2) was titrated with 0.050 M HCl. What is the pH of the solutionafter 10.00 ml of HCl is added?(a) 4.19(b) 5.09(c) 5.74(d) 6.2414. What is the ratio of moles of benzoate (C7H5O2‒) to benzoic acid (HC7H5O2) in the solution that results from thecombination of the NaC7H5O2 and HCl in the problem above?(a) 8(b) 0.125(c) 0.0040(d) 0.00050A buffer is prepared by adding 4.8 g of (NH4)2SO4 to 425 mL of 0.258 M NH3. Assuming that the volume stays constant, what is pH of the buffer solution? Consider: Kb (NH3) = 1.8×10–5, and Molar Mass of (NH4)2SO4 = 132.14 g/mol. (A) 10.04 (B) 5.22 (C) 9.44 (D) 4.93 (E) 1.75A chemist has synthesized a monoprotic weak acid and wants to determine its pKa value. To do so, they dissolved 2.00 mmol of the solid acid in 100.0 mL of water and then titrated the resulting solution with 0.0500 M NaOH. After20.0 mL of added NaOH, the pH of the solution was 6.00. What is the pKa of the acid?(a) 5.85 (b) 7.02 (c) 9.90 (d) 6.00
- Don’t ever mix acid with cyanide (CN) because it liberates poisonous HCN(g). But, just for fun, calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN with (a) 4.20 mL of 0.438 M HClO4. (b) 11.82 mL of 0.438 M HClO4. (c) What is the pH at the equivalence point with 0.438 M HClO4?A solution NH3 that contains 72 mL of 0.043 M ammonia, NH3, is titrated with 0.083 M HCl. The Kb of ammonia is 1.8x10-5. (a) What volume of 0.083 M HCI would be added to reach the equivalence point? Give the volume in mL. 49 37 mL (b) At the equivalence point, what is the pH of the solution? (Assume that volumes are additive.) 40 8.98 X whawhat "at onubralence point" implies about the quantities of the combined acids and bases? Did you remember(7) Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid( HC H O,,K=1.3x105) (b) 0.1000M sodium propanoate (Na C HỎ) (c) 0.1000M HC₂H₂O, and 0.1000M Nа С¸¸0₂ 3 5 52 (d) After 0.020 mol of HCl is added to 1.00 L solution of (a) and (b) above. (e) After 0.020 mol of NaOH is added to 1.00 L solution of (a) and (b) above.
- PR: (a) Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of sodium cyanate (NaCNO) and 1.0 mole of cyanic acid (HCNO) in enough water to make 800. mL of solution. (10). (b) Calculate the pH after 0.050 moles of NaOH are added to the buffer solution (assume no volume change) (10). Ka (HCNO) = 2.0 × 10-4A 500.-mL solution consists of 0.050 mol of solid NaOH and 0.13 mol of hypochlorous acid (HClO; Kₐ=3.0X10⁻⁸) dis-solved in water.(a) Aside from water, what is the concentration of each species that is present?(b) What is the pH of the solution?(c) What is the pH after adding 0.0050 mol of HCl to the flask?7. A 0.040 M solution of a monoprotic acid (HA) is 14% ionized. Calculate the ionization constant, Kc of the acid. 8. (a) Calculate the percent ionization of a 0.20 M solution of the monoprotic acetylsalicylic acid (aspirin) for which Ka = 3.0 x 10-4 .(b)The pH of gastric juice in the stomach of a certain individual is 1.00. After a few aspirin tablets have been swallowed, the concentration of acetylsalicylic acid in the stomach is 0.20M. Calculate the % ionization of the acid under these conditions. What effect does the nonionized acid have on the membranes lining of the stomach? 9. Calculate the [H+], [C2O4-2], and [H2C2O4] in a 0.0010M H2C2O4 solution. (Ka = 5.4 x 10-2)
- Which is the best solution to buffer at pH = 10.50? (A) A buffer solution containing 0.25 M H2CO3 and 0.45 M Na2CO3; (Ka (H2CO3) = 4.3 x 10–7) (B) A buffer solution containing 0.25 M NaHCO3 and 0.45 M Na2CO3; (Ka (HCO3–) = 5.6 x 10–11) (C) A buffer solution containing 0.12 M NaHCO3 and 0.22 M Na2CO3; (Ka (HCO3–) = 5.6 x 10–11) (D) A buffer solution containing 0.12 M NH4Cl and 2.2 M NH3; (Ka (NH4+) = 5.6 x 10–10)6. (a) Consider the pH of a solution containing a weak acid, HA , and a soluble salt of the weak acid, such as NaACalculate the pH change that results when 11 mL of 5.1 M NaOH is added to 790. mL of each the following solutions. Use the Acid-Base Table. (a) pure water 4.0 5.84 (b) 0.10 M NH4CI 4.0 5.43 (c) 0.10 M NH3 4.0✔ X (d) a solution that is 0.10 M in each NH4+ and NH3 4.0✔