A diprotic acid, H,A, has acid dissociation constants of Ka = 4.88 x 10-4 and K2 = 5.41 x 10-12. Calculate the pH and molar concentrations of H,A, HA", and A?- at equilibrium for each of the solutions. A 0.112 M solution of H, A. pH = [H,A] = M [HA-] = [A²-] = M A 0.112 M solution of NaHA. pH= |H,A] = M [HA"] = M [A²-] = M %3!

A diprotic acid, H,A, has acid dissociation constants of Ka = 4.88 x 10-4 and K2 = 5.41 x 10-12. Calculate the pH and molar concentrations of H,A, HA", and A?- at equilibrium for each of the solutions. A 0.112 M solution of H, A. pH = [H,A] = M [HA-] = [A²-] = M A 0.112 M solution of NaHA. pH= |H,A] = M [HA"] = M [A²-] = M %3!

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter13: Acids And Bases

Section: Chapter Questions

Problem 3RQ

See similar textbooks

Similar questions

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.
For conjugate acidbase pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq) where Ka = 1.8 105 a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to part b. why do we classify the acetate ion (CH3CO2) as a weak base? Use an appropriate reaction to justify your answer. In general, as base strength increases, conjugate acid strength decreases. Explain why the conjugate acid of the weak base NH3 is a weak acid. To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is 1 106, calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids. To tie it all together, some instructors have students think of Li+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ as the conjugate acids of the strong bases LiOH, KOH. RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not technically correct, the conjugate acid strength of these cations is similar to the conjugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the correct response. The conjugate base of a weak acid is a_____base. The conjugate acid of a weak base is a_____acid. The conjugate base of a strong acid is a_____base. The conjugate acid of a strong base is a_____ acid. (Hint: Weak gives you weak and strong gives you worthless.)
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?
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 strategies are also followed when solving for the pH of a base in water. What is the strategy for calculating the pH of a strong base in water? List the strong bases mentioned in the text that should be committed to memory. Why is calculating the pH of Ca(OH)2 solutions a little more difficult than calculating the pH of NaOH solutions? Most bases are weak bases. The presence of what element most commonly results in basic properties for an organic compound? What is present on this element in compounds that allows it to accept a proton? Table 13-3 and Appendix 5 of the text list Kb values for some weak bases. What strategy is used to solve for the pH of a weak base in water? What assumptions are made when solving for the pH of weak base solutions? If the 5% rule fails, how do you calculate the pH of a weak base in water?
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?
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
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
Using the diagrams shown in Problem 10-117, which of the solutions would have the greatest buffer capacity, that is, greatest protection against pH change, when the following occurs? a. A strong acid is added to the solution. b. A strong base is added to the solution.
A diprotic acid, H, A, has acid dissociation constants of Kal = 4.58 x 10-4 and K2 = 3.60 x 10-11. Calculate the pH and molar concentrations of H, A, HA", and A2- at equilibrium for each of the solutions. A 0.184 M solution of H, A. pH = [H,A] = M [HA-] = [A?-] = M M A 0.184 M solution of NaHA. pH= [H,A] = M [HA-] = [A2-] = M M A 0.184 M solution of Na, A. [H,A] = pH= M [HA"] = [A?-] = M M
Ascorbic acid (H, C,H,0,) is a diprotic acid. The acid dissocation constants for H,C,H,O, are Kal = 8.00 x 10-5 and K„2 = 1.60 x 10-12. Determine the pH of a 0.134 M solution of ascorbic acid. 2.49 pH = Determine the equilibrium concentrations of all species in the solution. [H,C,H,O,] = 0.131 M [HC,H,0G] = 3.22 x10-3 M 7.18 x10 -8 [C,H,O; = M Incorrect
Ascorbic acid (H,C, H,0,) is a diprotic acid. The acid dissocation constants for H,C. H,O, are K - 800 x 10 and Ka- 1.60 x 10-1 Determine the pH of a 0.138 M solution of ascorbic acid. pH = Determine the equilibrium concentrations of all species in the solution. (H,C,H,0, = M |HC, H,0;] - C,H,O-