Consider a titration of 33.8 mL of 1.64 M solution of aqueous ammonia [Kb(NH3) =
ID: 823165 • Letter: C
Question
Consider a titration of 33.8 mL of 1.64 M solution of aqueous ammonia [Kb(NH3) = 1.8x10^-5] with a 1.24 M solution of hydrochloric acid. Calculate:
a) The pH of the aqueous ammonia solution before the titration,
b) The pH of the solution at half-equivalence point,
c) The pH after 7.5 mL of HCL were added.
d) The pH of the solution at the equivalence point.
e) The pH of the solution when 1.00 mL of the HCl have been added after the equivalence point was reached.
If you could please show me each step I would greatly appreciate it.
Explanation / Answer
a) [OH-]=square root of Kb x Cb (initial concentration)
b) pOH=pKb
pOH= -log of Kb
subtract the pOH from 14 to get pH
d) first find the moles of NH3; moles= concentration x volume
NH3 +HCL ----> NH4Cl
find the volume of HCL: moles/concentration
add the volume of NH3 to the volume of HCL for your total volume
find the concentration of NH4Cl: moles/total volume
NH4Cl -----> NH4+ + Cl-
NH4+ will hydrolyze
NH4+ + H2O <------> NH3 + H30
do an ice table
find Ka: (Kw/Kb) Kw= 1.0 x 10^-14
Ka= x^2/concentration of NH4 (it's just the concentration of NH4Cl) - x
you can use the approximation
pH = -log (square root of Ka x the concentration)