Hydrazoic acid, HN3, and butyric acid, HC4H7O2, are both weakmonoprotic acids. T

Question

Hydrazoic acid, HN3, and butyric acid, HC4H7O2, are both weakmonoprotic acids. Their pKa values (25 Degree C) are 4.72 and 4.8respectively. What is ( HC4H7O2) in a solution in which thestoichiometric concentrations of HN3 and HC4H7O2 are 0.121 mol/Land 1.83 * 10^-6 mol/L respectively. Hint : Stoichiometric concentrationreffers to the total conc. in solution, i.e total amount of aciddissolved in solution. Hydrazoic acid, HN3, and butyric acid, HC4H7O2, are both weakmonoprotic acids. Their pKa values (25 Degree C) are 4.72 and 4.8respectively. What is ( HC4H7O2) in a solution in which thestoichiometric concentrations of HN3 and HC4H7O2 are 0.121 mol/Land 1.83 * 10^-6 mol/L respectively. Hint : Stoichiometric concentrationreffers to the total conc. in solution, i.e total amount of aciddissolved in solution.

Explanation / Answer

   Since the given HN3, Butyric acid HC4H7O2 bothare the monoprotic acids .when ever these two are mixed onewith    other the new generated H2O also as weak acid.Anyhow all of three HN3 is greater acitic than other two acids.                 So      Ka = 1.905*10-5 forHN3                          =[H+ ][N3-]/[HN3]                                  initial                                                                                                         Eqilibrium     Concentration(mol/L)                                      xmol/LHN3                               Concentration (mol/L)       [HN3]0=0.121                                     ------------------->                         [HN3]=0.121-x      [H+] =0                                                 Dissociates                                        [H+ ] =x       [N3- ]=0                                                                                                           [N3- ] =x                                          Ka= 1.905*10-5                                       = x*x/0.121-x                                       ˜x2 /0.121                                 x= 0.2305*10-5                             Sicne by using 5% rule                                   = (0.2305*10-5 /0.121)*100                                  =0.001%      it is less than5%                          [H+ ] = x= 0.2305*10-5           pH= -log [H+ ]               =5+0.637               =5.637          [HC4H7O2]=  [HC4H7O2]0   -amount of HC4H7O2dissociated ˜ [HC4H7O2]0 =1.83*10-6                                Ka= 1.58*10-5                                   Ka  =[H+ ] [C4H7O2- ] /[HC4H7O2]                                   1.58*10-5  = 0.2305*10-5 [C4H7O2- ] /1.83*10-6                            [C4H7O2- ]=(1.58*10-5)(1.83*10-6)/0.2305*10-5                            [C4H7O2- ]=(1.58*10-5)(1.83*10-6)/0.2305*10-5

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