Construct the expression for Kb for the weak base, N:H«. N:H.(aq) + H:O(1) = OH-(aq) + N:Hs*(aq) 1 Based on the definition of Kb, drag the tiles into the numerator or denominator to construct the expression for the given base. Ko 5 RESET [H:O] [H.O) [OH] [N:H] [N:Ha] [NaHs] 2[H:O] 2[H.O"] 2[OH] 2[N:Hc*] 2[N:H4] 2[NaHs] [H:O) [HO*P [OH] [N:H«JF

Construct the expression for Kb for the weak base, N:H«. N:H.(aq) + H:O(1) = OH-(aq) + N:Hs(aq) 1 Based on the definition of Kb, drag the tiles into the numerator or denominator to construct the expression for the given base. Ko 5 RESET [H:O] [H.O) [OH] [N:H] [N:Ha] [NaHs] 2[H:O] 2[H.O"] 2[OH] 2[N:Hc] 2[N:H4] 2[NaHs] [H:O) [HO*P [OH] [N:H«JF

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter14: Acids And Bases

Section: Chapter Questions

Problem 35Q: Consider the following mathematical expressions. a. [H+] = [HA]0 b. [H+] = (Ka [HA]0)1/2 c. [OH] =...

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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.
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Consider the following mathematical expressions. a. [H+] = [HA]0 b. [H+] = (Ka [HA]0)1/2 c. [OH] = 2[B]0 d. [OH] = (Kb [B]0)1/2 For each expression, give three solutions where the mathematical expression would give a good approximation for the [H+] or [OH]. [HA]0 and [B]0 represent initial concentrations of an acid or a base.
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Pure sulfuric acid also autoionizes. Write the ion-product constant expression, Ksulf, and find the concentration of the conjugate base at 20oC (Ksulf = 2.7x10-4 at 20oC).
Consider the following list when answering the questions that follow: C6H5CO2H , Ka = 6.3 x 10-5 C6H5CO2 - , Kb = 1.6 x 10-10 CN- , Kb = 2.5 x 10-5 HCN , Ka = 4.0 x 10-10 For the reaction between benzoic acid (C6H5CO2H ) and a solution containing cyanide ions (CN- ): (a) The name of the conjugate base produced is --- (b) The name of the conjugate acid produced is --- (c) The reaction is ----- favoured (type 'reactant' or 'product'), because the reactants are the ----- (type 'weaker' or 'stronger') set of acid and base.
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Give typed explanation A 0.10 M solution of an acid, HA, has a pH of 3.0. What is the Kb of the conjugate base, A-, of this acid? For Kb, I got 9.1 x 10 ^-12. I want me make sure my answer is correct. Pleacse show steps and write clearly. Thank you!
The value of Ka for acetic acid is 1.80×10-5.What is the value of Kb, for its conjugate base, CH3COO-?
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Please don't provide handwritten solution... In the laboratory, a general chemistry student measured the pH of a 0.513 M aqueous solution of triethanolamine, C6H15O3N to be 10.721. Use the information she obtained to determine the Kb for this base.