Consider titrating 15.0 mL of a 0.245 M solution of H_2NNH_2 with a 0.453 M HCl
ID: 527838 • Letter: C
Question
Consider titrating 15.0 mL of a 0.245 M solution of H_2NNH_2 with a 0.453 M HCl solution. The k_o value for H_2NNH_2 is 1.3 times 10^-4. a. Without doing any calculation, determine whether the pH of the solution at the equivalent point is greater than 7, equal to 7, or less than 7. b. What is the pH of the solution before any HCl has been added? c. What is the pH of the solution at the equivalent point? d. What is the pH of the solution when 5.0 mL more of the HCl solution is added after the equivalent point has been reached?Explanation / Answer
a) At equivalence point ,the pH will be less than 7
b) Kb =1.3×10^-4
H2N NH2 + H2O -------> N2H5+ + OH-
Kb = [ N2H5+] [ OH-]/[H2NNH2]
At equilibrium
[ N2H5+ ] = X
[ OH-] = X
[ H2NNH2 ] = 0.245M
Therefore,
1.3 × 10^-4 = X^2/(0.245 -X)
X^2 + 1.3×10^-4X - 3.19 × 10^-5 = 0
X = 5.58×10^-3
[OH-] = 5.58 × 10^-3M
pOH = 2.25
pH = 14 - 2.25
= 11.75
c) equivalence point = 8.11
Total volume = 15 + 8.11 = 23.11ml
At equivalence point
[ N2H5+ ] = 0.245M/ dilution = 0.245M/1.54 =0.1591M
N2H5+ + H2O -------> H3O+ + N2H4
Ka = [H3O+] [ N2H4]/[ N2H5 ]
Ka = Kw/Kb
Kw = 1×10-14
Kb = 1.3 ×10^-4
Ka = 1×10^-14/1.3×10^-4
= 7.7 × 10^-11
7.7×10^-11 = X^2/(0.159 - X)
X^2 + 7.7×10^-11X - 1.22 × 10^-11 = 0
X = 3.49×10^-6
[ H3O+ ] = 3.49 × 10^-6M
pH = 5.46
d) Total volume = 15ml + 8.11ml + 5 = 28.11ml
[ HCl ] = 0.453M/5.622 = 0.0806M
[ H3O+ ] = 0.0806M
pH = 1.09