Consider the titration of 100.0 mL of 0.100 M HONH 2 by 0.200 M HCl. For each vo
ID: 806806 • Letter: C
Question
Consider the titration of 100.0 mL of 0.100 M HONH2 by 0.200 M HCl.
For each volume of HCl added, decide which of the components is a major species after the HCl has reacted completely.
Kb for HONH2 = 1.1 x 10-8.
0.00 mL HCl added 40.00 mL HCl added 50.00 ml HCl added 100.00 mL HCl added
no: H+
yes: H2O
no: Cl-
yes: HONH2
no: HONH3
1.Calculate the pH at the equivalence point for this titration.
2. At what volume of HCl added in this titration does the pH = 6.04? Express your answer in mL and include the units in your answer.
Explanation / Answer
HONH2 + HCl ---> HONH3+ + Cl-
at the equivalence point
n acid = n base = M*V = 0.100*0.100 = 1.00*10^-2 mol
V acid added = n / M HCl = 1.00*10^-2 / 0.200 = 5.00*10^-2 L = 50.0 mL
total volume at EP = 0.100 L + 5.00*10^-2 L = 0.150 L
[HONH3+]= n / V = 1.00*10^-2 / 0.150 = 6.67*10^-2 M
the solution will be acidic
[H+] = square root (6.67*10^-2*Kw/Kb) = 2.46*10^-4 M
pH = 3.61
pKa of HONH3+ is 6.04
pH=pKa means half of the equivalence point that means V HCl = 25.0 mL