For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH.

Question

For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb = 1.8 x 10-5.

(a) Before the addition of any HCl solution. pH = the tolerance is +/-1 in the 4th significant digit

(b) After 20. mL of the acid has been added. pH = the tolerance is +/-1 in the 3rd significant digit

(c) After half of the NH3 has been neutralized. pH = the tolerance is +/-1 in the 3rd significant digit

(d) At the equivalence point. pH = the tolerance is +/-1 in the 3rd significant digit

Explanation / Answer

(a) Before addition of HCl

NH3 + H2O <==> NH4+ + OH-

let x amount has reacted,

Kb = 1.8 x 10^-5 = x^2/0.1

x = [OH-] = 1.34 x 10^-3 M

pOH = -log[OH-] = 2.87

pH = 14 - pOH = 11.13

(b) after 20 ml of 0.1 M HCl is added

moles of NH3 = 0.1 M x 0.05 L = 5 x 10^-3 mols

moles of HCl = 0.1 M x 0.02 L = 2 x 10^-3 mols

excess moles of NH3 = 3 x 10^-3 mols

molarity of NH4+ = 2 x 10^-3/0.07 = 0.0286 M

molarity of NH3 = 3 x 10^-3/0.07 = 0.043 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = Kw/Kb = 1 x 10^-14/1.8 x 10^-5 = 5.56 x 10^-10

pKa = 9.255

pH = 9.255 + log(0.043/0.0286)

      = 9.432

(c) after half of NH3 has been neutralized.

This is half equivalence point

molarity of NH4+ = molarity of NH3

So, log([NH3]/[NH4+]) = 0

pH = pKa = 9.255

(d) At the equivalence point

moles of acid = moles of base

moles of NH3 = 0.1 M x 0.05 L = 5 x 10^-3 mols

So moles of HCl = 5 x 10^-3 mols

Volume of HCl added = 5 x 10^-3/0.1 = 0.05 L

molarity of NH4+ = 5 x 10^-3/0.1 = 0.05 M

NH4+ + H2O <==> NH3 + H3O+

let x amount has reacted,

Ka = [NH3][H3O+]/[NH4+]

5.56 x 10^-10 = x^2/0.05

x = [H+] = 5.27 x 10^-6 M

pH = -log[H+] = 5.28

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