Question

Find the pH at the equivalence point(s) and the volume (mL) of 0.125 M HCl needed to reach the point(s) in titrations of the following. (See this table.) (a) 69.0 mL of 0.225 M NH3 volume to reach equivalence point pH at equivalence point

Answer 1

NH3 + HCl = > NH4 + + Cl-

at the equivalence point NH3 has reacted with the HCl and in solution we have the salt NH4+Cl-

moles of NH3 = = .225 x 69 / 1000 = 0.0155moles = moles of HCl = moles of NH4 +

volume of HCl = moles / M = .0155 / .125 = 0.124 L or 124 mL;

volume of HCl needed = 124 mL

total volume = 124 + 69 = 193 mL

the concentration of NH4+ = .0155 moles in 193 mL or 0.0155 x 1000 / 193 M =0.0803 M
the HN4+ at the end point is hydrolyzed in the reaction
NH4+ + H2O = > NH3 + H+
Ka = [NH3] [H+ ] / [NH4+}
Ka (NH4+) is derived from Kb(NH3)
Ka x Kb = 10 ^ -14 ; Ka = 10 ^ -14 / 1.76 10 ^ -5 = 5.68 x 10 ^-10

if x moles of NH4+ dissociate we have at equilibrium [ H+ ] = [NH3 ] = x
[NH4] = .0803 -x

Ka = [NH3] [H+ ] / [NH4+]
5.68 x 10 ^-10 = x ^2 / .0803-x
x is very small so the solution simplifies to


x= 6.75 x 10 ^ -6
[H+] = 6.75 x 10 ^ -6

pH at the equivalence point = 5.17