Suppose you dissolve 1 mole of benzoic acid (C_6H_5COOH, pK_a = 4.2) and 1 mole

Question

Suppose you dissolve 1 mole of benzoic acid (C_6H_5COOH, pK_a = 4.2) and 1 mole of tri-methylamine, (CH_3)_3 N, in a liter of water. The trimethylammonium ion, (CH_3)_3 NH^+, has a pKa of 9.8. Answer the following questions (please leave the answers to parts A-C in exponent form): What will be the numerical value of the equilibrium constant [C_6 H_5 COO^-][(CH_3)_3 NH^+]/[C_6H_5 COOH][(CH_3_3 N] ? Keeping in mind the stoichiometry of the Breasted acid-base reaction, what will be the numerical value of [(CH_3)_3 NH^+]/[(CH_3)_3N]? What must be the hydrogen ion concentration at equilibrium? What is the pH of the solution?

Explanation / Answer

A. Ka of benzoic acid/Ka of trimethylammonium we get {[C6H5COO-][(CH3)3NH+]}/{[C6H5COOH][(CH3)3N]}=10^-4.2*10^9.8=10^5.6

B. numerical value of [(CH3)3NH+]/[(CH3)3N]=1

C. [H+]=root over(Ka*C) of benzoic acid=root over(10^-4.2*1)=10^-2(nearly)

PH=-log[H+]=-log(10^-2)=2 (nearly)

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