Question

Calculate the pH at the equivalence point when 25.0mL of 0.100M NH3 is titrated with 0.100M HCl.

The reaction is given as follows :

NH3 + H30+ -> NH4+ + H20

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Answer 1

Ans:

Volume of NH₃ taken = 25.0 mL

Molarity of NH₃ taken = 0.100 M

No. of millimoles of NH₃ taken = 25.0mL x 0.100 M = 2.5 mmol

Molarity of HCl taken = 0.100 M

Reaction:

1 NH₃(aq) + 1 HCl(aq) --> 1 NH₄Cl (aq) (can be written as NH₃ + H₃0⁺ -> NH₄⁺ + H₂0)

For the reaction to reach completion 1 mol of NH3 demands 1 mol HCl which gives 1 mol of NH4Cl salt.

So, 2.5 mmol of NH3 demands 2.5 mmol of HCl which will form 2.5 mmol of NH4Cl salt.

Volume of 0.100 M HCl taken to reach equivalence point = 2.5mmol/0.100M = 25 mL    {because M = mol/vol(L)}

Total volume of the solution after reaching equivalence point = 25mLNH₃ + 25mLHCl = 50 mL = V

Molarity of NH4Cl after reaching equivalence point = mol/volume = 2.5mmol/50mL = 0.05 M (C)

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pKb of NH3 = 4.75 (data collected)

Molarity of NH4Cl, C = 0.05

NH4Cl is a salt made up of weak base and strong acid, whose pH is given by

pH = 7-{(pKb+logC)/2}

pH = 7-{(4.75 + log0.05)/2} = 5.2755

pH at the equivalence point = 5.2755