Question

20.00 cm³ of 0.1 M ammonia solution is titrated with 0.25 M HClO₄. What is the added volume of titrant and the pH at 75% degree of titration?

Answer 1

The molar amount of ammonia: n = c × V = 0.1 M × 0.020 L = 0.002 mol

At 100 % titration, nHClO₄ = nNH₃ = 0.002 mol At 

75 % titration, nHClO₄ = 0.75 × n_NH3 

= 0.0015 mol HClO₄. 

We know the concentration (0.25 M) of the HClO₄ solution, so, the volume can be calculated: 

At 75 % titration, we have a buffer, NH₃ and NH₄Cl together. 75 % of the original NH₃ amount is converted to NH₄Cl, and 25 % is remained as NH₃. So, using the equation for buffers: