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The hydronium ion concentration of an aqueous solution of 0.441 M acetylsalicylic acid (aspirin), HC9H7O4, (Ka = 3.00×10-4) is
[H3O+] =
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- Acrylic acid is used in the polymer industry in the production of acrylates. Its K, is 5.6 X 10“’. What is the pH of a 0.11 M solution of acrylic acid, CH2CHCOOH?COHSOH(ag) + H2On + CeHsO (aq) + H3O*(a9) Ka= 1.12 x 10-10 (a) Phenol is a weak acid that partially dissociates in water according to the equation above. Write the equilibrium-constant expression for the dissociation of the acid in water. (b) What is the pH of a 0.75 M CaHsOH(ag) solution? (C) For a certain reaction involving CaHsOH(ag) to proceed at a significant rate, the phenol must be primarily in its deprotonated form, C3H5O (eg). In order to ensure that the CsHsOH(aq) is deprotonated, the reaction must be conducted in a buffered solution. On the number scale below, circle each pH for which more than 50 percent of the phenol molecules are in the deprotonated form (CoHsO (aq). Justify your answer. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Justification: (d) CeHsOH(ag) reacts with NaOH(ag). Write a net ionic equation representing this reaction (aka: invasion equation). (e) What is the pH of the resulting solution when 30 mL of 0.40 M CSH5OH(aq) is added to 25 mL of 0.60 M NAOH. Show all work…Propionic acid, HC3H5O2, has Ka= 1.34 x 10–5. (a) What is the molar concentration of H3O+ in 0.15 M HC3H5O2 and the pH of the solution? (b) What is the Kb value for the propionate ion, C3H5O2–? (c) Calculate the pH of 0.15 M solution of sodium propionate, NaC3H5O2. (d) Calculate the pH of solution that contains 0.12 M HC3H5O2 and 0.25 M NaC3H5O2.
- The major component of vinegar is acetic acid, CH3COOH. Its Ka is 1.8 × 10-5 . One student used 1.000 M NaOH to titrate 25.00 mL vinegar. At the end point, 21.82 mL NaOH was used. (a) What is the concentration of CH3COOH in vinegar? (b) What is the pH of the solution at the end point? (c) What indicator(s) the student should use in this titration? Explain8. (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)Calculate the pH and the pOH of these solutions: (c) 0.957 M HC2H3O2 (Ka = 1.8 x 10-5) pH = pOH =
- (a) Calculate the percent ionization of a 0.20 M solution of monoprotic acetylsalicylic acid (aspirin) for which Ka = 3.0 x 10^-4. (b) The pH of gastric acid 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 0.20 M. Calculate the percent ionization of the acid under these conditions.A student is provided with a 0.1 M stock solution of NaOH. Student was then asked to mix 50.0 mL of this stock solution of NaOH with 450.0 mL of water to prepare a dilute solution of NaOH. (A) Calculate the molarity of the diluted solution. (B). Calculate the hydronium ion concentration, [H3O+] in the final diluted solution. (8 points) (C ) Calculate the pH of the diluted NaOH solution. (D) Is the final diluted solution acidic, basic or neutral?The pH of an acetic acid (Ka = 1.75 x 10−5) solution is 3.14. What is the concentration of acetic acid in molarity?
- What is the hydronium-ion concentration of a 1.76 x 104 M solution of p-bromobenzoic acid, BrCgH4 COOH, for which Ka = 1.00 x 10-42 [H3O+]=| |мThe pH of an aqueous solution of 0.400 M hydrocyanic acid, (Ka (HCN) = 4.00 × 10-10) isDetermine the pH of (a) A 0.10 M CH3COOH solution (ICE) (b) A solution that is 0.10 M CH3COOH and 0.10 M CH3COONa. Ka for CH3COOH = 1.8 X 10-5.
