millimols HCl = 34 x 0.2 = 6.8
mmols NH3 = 29 x 0.05 = 1.45
..........HCl + NH3 ==> NH4Cl
I.........6.8...1.45......0
C........-1.45.-1.45.....+1.45
E.........5.35...0........1.45
If you look at the numbers you will notice that the titration with HCl has neutralized all of the NH3 and the point in the titration is well above the equivalence point. So you have excess HCl in NH4Cl.
(H^+) = excess HCl and pH from that.
(HCl) = mmols/mL = 5.35/63 = ?
What is the pH at the point during a titration when 34 mL of 0.200 M HCl has been added to 29 mL of 0.0500 M NH3 solution? HCl is a strong acid, Kb=1.8×10−5 for NH3.
Answer in units of pH
1 answer