1 3. 10 11 12 13 14 Calculate the pH at 25 °C of a 0.69M solution of potassium acetate (KCH,CO,). Note that acetic acid (HCH,CO,) is a weak acid with a p K, of 4.76. Round your answer to 1 decimal place. alb pH =

1 3. 10 11 12 13 14 Calculate the pH at 25 °C of a 0.69M solution of potassium acetate (KCH,CO,). Note that acetic acid (HCH,CO,) is a weak acid with a p K, of 4.76. Round your answer to 1 decimal place. alb pH =

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter9: Acids, Bases, And Salts

Section: Chapter Questions

Problem 9.108E

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Table 13-4 lists the stepwise Ka values for some polyprotic acids. What is the difference between a monoprotic acid, a diprotic acid, and a triprotic acid? Most polyprotic acids are weak acids; the major exception is H2SO4. To solve for the pH of a solution of H2SO4, you must generally solve a strong acid problem as well as a weak acid problem. Explain. Write out the reactions that refer to Ka1 and Ka2 for H2SO4. For H3PO4, Ka1 = 7.5 103, Ka2 = 6.2 108, and Ka3= 4.8 1013. Write out the reactions that refer to the Ka1, Ka2and Ka3equilibrium constants. What are the three acids in a solution of H3PO4? Which acid is strongest? What are the three conjugate bases in a solution of H3PO4? Which conjugate base is strongest? Summarize the strategy for calculating the pH of a polyprotic acid in water.
Amino acids are an important group of compounds. At low pH, both the carboxylic acid group (CO2H) and the amine group (NHR) are protonated. However, as the pH of the solution increases (say, by adding base), the carboxylic acid proton is removed, usually at a pH between 2 and 3. In a middle range of pHs, therefore, the amine group is protonated, but the carboxylic acid group has lost the proton. (This is called a zwitterion.) At more basic pH values, the amine proton is dissociated. What is the pH of a 0.20 M solution of alanine hydrochloride, [NH3CHCH3CO2H]Cl?
Malic acid is a weak diprotic organic acid with Ka1 = 4.0 104 and Ka2 = 9.0 105. a Letting the symbol H2A represent malic acid, write the chemical equations that represent Ka1 and Ka2. Write the chemical equation that represents Ka1 Ka2. b Qualitatively describe the relative concentrations of H2A, HA, A2, and H3O+ in a solution that is about one molar in malic acid. c Calculate the pH of a 0.0175 M malic acid solution and the equilibrium concentration of [H2A]. d What is the A2 concentrationin in solutions b and c?
Estimate the pH that results when the following two solutions are mixed. a) 50 mL of 0.3 M CH3COOH and 50 mL of 0.4 M KOH b) 100 mL of 0.3 M CH3COOH and 50 mL of 0.4 M NaOH c) 150 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2 d) 200 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2
Composition diagrams, commonly known as alpha plots, are often used to visualize the species in a solution of an acid or base as the pH is varied. The diagram for 0.100 M acetic acid is shown here. The plot shows how the fraction [alpha ()] of acetic acid in solution, =[CH3CO2H][CH3CO2H]+[CH3CO2] changes as the pH increases (blue curve). (The red curve shows how the fraction of acetate ion, CH3CO2, changes as the pH increases.) Alpha plots are another way of viewing the relative concentrations of acetic acid and acetate ion as a strong base is added to a solution of acetic acid in the course of a titration. (a) Explain why the fraction of acetic acid declines and that of acetate ion increases as the pH increases. (b) Which species predominates at a pH of 4, acetic acid or acetate ion? What is the situation at a pH of 6? (c) Consider the point where the two lines cross. The fraction of acetic acid in the solution is 0.5, and so is that of acetate ion. That is, the solution is half acid and half conjugate base; their concentrations are equal. At this point, the graph shows the pH is 4.74. Explain why the pH at this point is 4 74.
Two students were asked to determine the Kb of an unknown base. They were given a bottle with a solution in it. The bottle was labeled aqueous solution of a monoprotic strong acid. They were also given a pH meter, a buret, and an appropriate indicator. They reported the following data: volume of acid required for neutralization = 21.0 mL pH after 7.00 mL of strong acid added = 8.95 Use the students' data to determme the Kb of the unknown base.
A solution has a pH of 4.5. What would be the color of the solution if each of the following indicators were added? (See Fig. 14-8.) a. methyl orange b. alizarin c. bromcresol green d. phenolphthalein
Calculate the pH of each of the following solutions. (a) 10.0 mL of 0.300 M hydrofluoric acid plus 30.0 mL of 0.100 M sodium hydroxide (b) 100.0 mL of 0.250 M ammonia plus 50.0 mL of 0.100 M hydrochloric acid (c) 25.0 mL of 0.200 M sulfuric acid plus 50.0 mL of 0.400 M sodium hydroxide
The pigment cyanidin aglycone is one of the anthocyanin molecules that gives red cabbage (Brassica oleracea var. capitata f. rubra) its characteristic red coloration. Many chemistry students have used this red cabbage indicator to study acid-base chemistry. Estimate tire pH range at which cyanidin agly-cone shows a color change. Anth-H(aq) Anth(aq) + H+ (aq) Ka = 1.3 107
Develop a set of rules by which you could predict the pH for solutions of strong or weak acids and strong or weak bases without using a calculator. Your predictions need to be accurate to 1 pH unit. Assume that you know the concentration of the acid or base and that for the weak acids and bases you can look up the pKa (log Ka) or Ka values. What rules would work to predict pH?
Methyl orange, HMO, is a common acid-base indicator. In solution it ionizes according to the equation: HMOaqH+aq+MO-aqredyellow If methyl orange is added to distilled water, the solution turns yellow. If 1 drop or two of 6 M HCl is added to the yellow solution, it turns red. If to that solution one adds a few drops of 6 M NaOH, the color reverts to yellow. a. Why does adding 6 M HCl to the yellow solution of methyl orange tend to cause the color to change to red? Note that in solution HCl exists as H+ and Cl- ions. b. Why does adding 6 M NaOH to the red solution tend to make it turn back to yellow? Note that in solution NaOH exists as Na+ and OH- ions. How does increasing OH- shift Reaction 3 in the discussion section? How would the resulting change in H+ affect the dissociation reaction of HMO?