For the titration of 65.0 mL of 0.250 M NH3 with 0.500 M HCl at 25°C, determine
ID: 812390 • Letter: F
Question
For the titration of 65.0 mL of 0.250 M NH3 with 0.500 M HCl at 25°C, determine the relative pH at each of these points:
(a) before the addition of any HCl, is pH (> or = or <) 7?
(b) after 32.5 mL of HCl has been added, is pH (> or = or <) 7?
(c) after 52.5 mL of HCl has been added, is pH (> or = or <) 7?
Calculate the pH for each of the following cases in the titration of 50 mL of 0.160 M HClO(aq) with 0.160 M KOH(aq). The ionization constant for HClO is 4.0×10^-8.
(d) before addition of any KOH
pH=
(e) after addition of 25.0 mL of KOH
pH=
(f) after addition of 40.0 mL of KOH
pH=
(g) after addition of 50.0 mL of KOH
pH=
(h) after addition of 60.0 mL of KOH
pH=
Explanation / Answer
???? a) pH must be more than 7 since it is a base, OH? ions are higher thn H+
b) Calculate amount of HCl vs NH3
V1*M1 = 65*0.25 = 16.25 mmol of NH3
V2*M2 = 0.5*32.5 = 16.25 mmol of HCl
You will neutralize all base. In equilibrium you will see:
NH3 + H2O <?> NH4OH + H+
Expect an acidic value... pH < 7
c) After 52.5 ml of HCl you will definitely see acidic solution (more HCl than NH3) Since HCl is a strong
acid it will be acidic
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