Assume you titrate 20.0 mL of 0.11 M NH 3 with 0.10 M HCl. (a) What is the pH of

Question

Assume you titrate 20.0 mL of 0.11 M NH3 with 0.10 M HCl.

            (a) What is the pH of the NH3 solution before the titration begins?

            (b) What is the pH of the equivalence point?

            (c) What is the pH at the midpoint of the titration?

            (d) Which indicator in figure 15.8 on page 756 would be best to detect the

      equivalence point?

            (e) Calculate the pH of the solution after adding 5.00, 11.0, 15.0, 20.0,

      22.0, and 25.0 mL of the acid. Combine this information with that from

  (a) through (c) and plot the titration curve.

I mostly need help with part e

Explanation / Answer

NH3 + H2O <-----> NH4+ + OH-
Kb = 1.8 x 10^-5
1.8 x 10^-5 = x^2 / 0.11 -x
x = [OH-] =0.0014 M
pOH = - log 0.0014 = 2.9
pH = 14 - 2.9 =11.1


Moles NH3 = 0.0200 L x 0.11 = 0.0022
Moles HCl needed = 0.0022
V (HCl) = 0.0022 / 0.10 = 0.022 L
Total volume = 0.042 L
NH3 + H+ >> NH4+
NH4+ + H2O <----> NH3 + H3O+
K = Kw/ Kb = 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10
[NH4+] = 0.0022 / 0.042 = 0.052 M
5.6 x 10^-10 = x^2 / 0.052-x
x = [H3O+] = 5.4 x 10^-6 M
pH = 5.3

At the midpoint moles HCl needed = 0.0022 /2 = 0.0011
V ( HCl) = 0.0011 / 0.10 = 0.011 L
Total volume = 0.031 L
NH3 + H+ >> NH4+
Moles NH3 = 0.0022 - 0.0011 = 0.0011
Moles NH4+ = 0.0011
[NH3] = [NH4+] = 0.0011 / 0.031 =0.035
pOH = pKb + log [NH4+] / [NH3]
pKb = - log 1.8 x 10^-5 = 4.7
pOH = 4.7 + log 0.035 / 0.035 = 4.7

no figure

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