Calculate the pH and concentrations of H, A, HA¯, and A?-, at equilibrium for a 0.214 M solution of Na, A. The acid dissociation constants for H, A are Kal = 6.92 × 10-5 and K2 = 2.53 x 10-9. pH = [A?-] = M [HA¯] = М [H, A] = M

Calculate the pH and concentrations of H, A, HA¯, and A?-, at equilibrium for a 0.214 M solution of Na, A. The acid dissociation constants for H, A are Kal = 6.92 × 10-5 and K2 = 2.53 x 10-9. pH = [A?-] = M [HA¯] = М [H, A] = M

Introductory Chemistry: A Foundation

8th Edition

ISBN:9781285199030

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter16: Acids And Bases

Section: Chapter Questions

Problem 42QAP

See similar textbooks

Similar questions

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.)
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?
Weak base B has a pKb of 6.78 and weak acid HA has a pKa of 5.12. a Which is the stronger base, B or A? b Which is the stronger acid, HA or BH+? c Consider the following reaction: B(aq)+HA(aq)BH+(aq)+A(aq) Based on the information about the acid/base strengths for the species in this reaction, is this reaction favored to proceed more to the right or more to the left? Why? d An aqueous solution is made in which the concentration of weak base B is one half the concentration of its acidic salt, BHCl, where BH+ is the conjugate weak add of B. Calculate the pH of the solution. e An aqueous solution is made in which the concentration of weak acid HA twice the concentration of the sodium salt of the weak acid, NaA. Calculate the pH of the solution. f Assume the conjugate pairs B/BH+ and HA/A are capable of being used as color-based end point indicators in acidbase titrations, where B is the base form indicator and BH is the acid form indicator, and HA is the acid form indicator and A is the base form indicator. Select the indicator pair that would be best to use in each of the following titrations: (1) Titration of a strong acid with a strong base. (i) B/BH+ (ii) HA/A (2) Titration of a weak base with a strong acid. (i) B/BH+ (ii) HA/A
8-45 Describe how you would prepare each of the following solutions (in each case, assume that the base is a solid). (a) 400.0 mL of 0.75 M NaOH (b) 1.0 L of 0.071 M Ba(OH)2 (c) 500.0 mL of 0.1 M KOH (d) 2.0 L of 0.3 M sodium acetate
12.62 Write the formula of the conjugate acid of each of the following bases, (a) OH-, (b) NHj, (c) CHjNHt, (d) HPO/-, (e) CO.,2’
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?
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
A diprotic acid, H, A, has acid dissociation constants of Kaj = 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. [H,A] = pH = M [HA-] = [A?-] = M M A 0.112 M solution of NaHA. pH= (H,A] = M [HA"] = [A²-] = M M A 0.112 M solution of Na, A. pH= [H,A] = M [HA] = [A?-] = M M
A diprotic acid, H₂A, has acid dissociation constants of K₁1 = 3.75 x 104 and K₁2 = 5.10 × 10-¹¹. Calculate the pH and molar concentrations of H₂A, HA, and A²- at equilibrium for each of the solutions. A 0.195 M solution of H₂ A. pH = 2.08 -3 [HA] = 8.37 x107 A 0.195 M solution of NaHA. pH= 6.86 [HA] = pH= A 0.195 M solution of Na, A. Incorrect 11.78 [HA] = 6.09 X10-3 M M M [H₂A] = 0.187 [A²-] = [H₂A] = [A²-] = [H₂A] = 5.10 x10-11 6.76 x10-5 Incorrect 7.65 x10-5 Incorrect 2.67 X10-11 [A²] = 0.189 M M M M M M
A diprotic acid, H, A, has acid dissociation constants of Kal 4.76 x 10-4 and Ka2 5.22 x 10-12. Calculate the pH and molar concentrations of H, A, HA¯, and A2- at equilibrium for each of the solutions. A 0.124 M solution of H, A. [H,A] = pH = M [HA¯] = [A²-] = М M A 0.124 M solution of NaHA. pH= [H,A] = M [HA¯] = [A²-] = M M A 0.124 M solution of Na, A. pH= [H,A] = M [HA¯] = M [A²-] = M
A diprotic acid, H, A, has acid dissociation constants of K₁1 = 4.76 x 10-4 and K₁2 = 3.59 x 10-12. Calculate the pH and molar concentrations of H₂A, HA, and A2- at equilibrium for each of the solutions. A 0.205 M solution of H₂A. pH = [HA] = A 0.205 M solution of NaHA. pH= [HA] = M M [H₂A] = [A²-] = [H₂A] = [A²-] = M M M M
Ascorbic acid (H,C,H,O,) is a diprotic acid. The acid dissocation constants for H,C,H,O, are Kal = 8.00 × 10-5 and Ka2 = 1.60 x 10-12. Determine the pH of a 0.134 M solution of ascorbic acid. pH = Determine the equilibrium concentrations of all species in the solution. [H,C,H,O] = M [HC,H,0] = M [C,H,0;-] = M