Given the following two solutions: 0.200 MHC 2 H 3 O 2 (aq) with pKa = 4.76 0.20

Question

Given the following two solutions:    0.200 MHC2H3O2(aq) with pKa = 4.76    0.200 M NaOH(aq) A mixture is made using 50.0 mL of theHC2H3O2 and 25.0 mL of the NaOH.What is its pH? Please EXPLAIN youranswer. Given the following two solutions:    0.200 MHC2H3O2(aq) with pKa = 4.76    0.200 M NaOH(aq) A mixture is made using 50.0 mL of theHC2H3O2 and 25.0 mL of the NaOH.What is its pH? Please EXPLAIN youranswer.

Explanation / Answer

moles of HC2H3O2: 50 mL *(1L/1000mL)*(0.2 mol/L) = 0.01moles HC2H3O2 moles NaOH:   25 mL *(1 L/1000mL)*0.20 mol/L =0.005 moles OH-                         HC2H3O2    + OH-          C2H3O2-         +     H2O initial               0.01           0.005                     0 change            -0.005         -0.005                 +0.005                                   (acid base reactions go to completion) equil               0.005             0                         0.005 HC2H3O2 is also, at the end, in equil with C2H3O2- [HC2H3O2] = 0.005 moles/ 75 mL *(1000mL/1L) = 0.06667 M [C2H3O2-] = 0.005 mol/75 mL *(1000mL/1L) = 0.06667 M                       HC2H3O2           C2H3O2-         +     H+ initial              0.06667M                      0.06667M              0 change                  -x                                 +x                     +x equil            0.06667-x                       0.06667+x              x Ka = [H+][C2H3O2-]/[HC2H3O2] = (x)*(0.06667 + x)/(0.06667-x) =10-pKa = 10-4.76 = 1.7378e-5 assume 0.06667 >> x, so 0.06667 +x ~0.06667; 0.06667 -x ~ 0.06667 (x)*(0.06667 + x)/(0.06667-x) ~ x*(0.06667)/(0.06667) = 1.7378e-5 x = 1.7378e-5 = [H+] pH = -log[H+] = -log(1.7378e-5) = 4.76

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