Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: ?

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Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added:

(a) 0 mL,

 (b) 20.0 mL,

(c) 59.0 mL,

(d) 60.0 mL,

 (e) 61.0 mL,

(f) 65.0 mL.

1 Answer

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  • david
    Lv 7
    4 weeks ago

    1st -- you need kb for NH3 ,, maybe listed in your book as NH4OH

       .....  if I remember ,, it is 1.8x10^-5  ... if not you need to do this on your own

       a.   NH3  +  H2O  <-->  NH4+  +  OH-

             . . . . . . . . . . NH3 . . . NH4+ . . . OH-

              initial . . . . 0.050 . . .   0  . . . .  .  0

              change . . .   - x . . . .  x.  . . . . .   x

              equilib. . . .(0.050-x) . . x . . . . . . x

         Kb = [NH4+][OH-] / [NH3]

       1.8x10^-5  =  x^2 / (0.050)     ///  NH3 conc is rounded assuming x is small

       x^2 =  9x10^-7

       x =  0.000949 M = [OH-]

      pOH  =  3.02

      pH = 10.98

    =====================================

    so you do the rest

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