Many common weak bases are derivatives of NH3: where one or more of the hydrogen

Question

Many common weak bases are derivatives of NH3: where one or more of the hydrogen atoms have been replaced by another substituent. Such reactions can be genetically symbolized as where NX3 is the base and HNX3 + is the conjugate acid. The equilibrium-constant expression for this reaction is where Kb is the base ionization constant. The extent of ionisation, and thus the strength of the base, increases as the value of Kb increases. If Kb for NX3 is 9 times 10-6 what is the pOH of a 0.175 M aqueous solution of NX3? Express your answer numerically. If Kb for NX3 is 9 times 10-6, what is the percent ionization of a 0.325 M aqueous solution of NX3? Express your answer numerically. If Kb for NX3 is 9 times 10-6, what is the the pKa for the following reaction? Express your answer numerically.

Explanation / Answer

for part A. in order to find POH you need to know [OH-]. usethe following equation to find [OH-]. kb = [HNX3+][OH-]/[NX3] [NX3] = 0.175M, kb = 9 x 10-6, [OH-] =[HNX3+] so you can write the above equationas 9 x 10-6 = [OH-]2/0.175 after you find [OH-] plug it in to this equationpOH = -log[OH-] to find pOH. PART B find [OH-] first. 9 x 10-6 = [OH-]2/0.325 after you find it plug it into the following equation to fine% ionization. % ionization = [OH-]/[0.325] x 100% PART C 9 x 10-6 = [OH-]2/0.325 after you find it plug it into the following equation to fine% ionization. % ionization = [OH-]/[0.325] x 100% PART C ka x kb = 10-14 you know kb. find ka from the equation above. next plug it into the following equation Pka = -log[Ka] to find Pka. hope that helps you know kb. find ka from the equation above. next plug it into the following equation Pka = -log[Ka] to find Pka. hope that helps

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