Erik asked in Science & MathematicsChemistry · 4 weeks ago

# 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: ?

17.46

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.

Relevance
• 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