Question

Chapter 15, Question 67 Parameterization X Incorrect. Calculate the pH of a 0.55 M solution of aqueous C6H5NH3Br. X 2.45 the tolerance is +/-2% Click if you would like to show Work for this question: Open Show Work

Answer 1

Write the reaction for hydrolysis of CH_NH_Br as follows: CH, NH,Br CH NH,+ Br- Given, concentration of CH_NH_Br is 0.55 M [CH, NH,Br] = 0.55 M [CH, NH,4 ] = [CH,NH,Br] [CH, NH,] = 0.55 M Write the reaction for hydrolysis of CH, NH, as follows: CH,NH, * +H2O → CHÁNH, + Hạnh K for CH_NH, = 4.4x104 ку 1x10-14 = 2.273x10-11 K 4.4x10+ Draw the ICE table as follows: CHÚNH,* +H,O • CHÚNH + Hạnh 1(M): 0.55 0 0 C(M) -X E(M): 0.55-x x X X x [CH NH][Hoº] [CH,NH,*] Substitute the corresponding valuesin the above equation and calculate x. x2 2.273x10-11 = 0.55-x x = 0.55x2.273x10-11 x= V1.25x 10-11 x= 3.54x10 [H,0*]= 3.54x10-6 M Calculate the pH as follows: pH = -log[H3O+] pH = -log(3.54x100) pH = 5.45 Therefore, the pH of the solution is 5.45