Answer :
The pH after the following volumes of titrant have been added 0 ml 20 mL 59. 1 mL 60. 0 mL 71. 4 mL 73. 4 mL are 11.89, 11.89, 8.45, 8.45, 7.98, 8.95 respectively.
The reaction between NH3 and HCl can be represented by the following equation: NH3 + HCl → NH4+ + Cl-
To calculate the pH after different volumes of titrant have been added, we need to determine the amount of titrant that has reacted with the analyte and the resulting concentration of the products.
A. 0 mL of titrant (initial state)
At the start, there is no titrant added to the analyte, so the concentration of NH3 is 0.050 M. NH3 is a weak base, so we can use the Kb expression to calculate the concentration of OH-:
[tex]Kb = [NH4+][OH-] / [NH3][/tex]
[tex]1.8 * 10^{-5} = x^2 / (0.050 - x)[/tex]
initial concentration of NH3 is much greater than the initial concentration of HCl, we can assume that the concentration of NH3 does not change significantly during the titration.
[tex]Kb = x^2 / 0.050\\x = \sqrt{Kb * 0.050} = 1.3 * 10^{-3} M[/tex]
The concentration of OH- is equal to [tex]1.3 * 10^{-3} M[/tex], so we can calculate the pH:
[tex]pH = 14 - pOH = 14 - (-log[OH-]) = 11.89[/tex]
Therefore, the pH at the start of the titration is 11.89.
B. 20 mL of titrant
After adding 20 mL of 0.025 M HCl, the volume of the solution is 50 mL (30 mL NH3 + 20 mL HCl). The moles of HCl added is:
moles of HCl = volume x concentration = 0.020 L x 0.025 mol/L = 5 x 10^-4 mol
Since the reaction is a 1:1 reaction, the moles of NH3 remaining is equal to the moles of HCl added.
concentration of NH3 = moles of NH3 / volume of NH3 = (0.050 mol/L x 0.030 L - 5 x 10^-4 mol) / 0.030 L = 0.048 mol/L
Since the concentration of NH3 has decreased, we need to recalculate the concentration of OH- using the new concentration of NH3:
[tex]Kb = [NH4+][OH-] / [NH3]\\1.8 * 10^{-5} = x^2 / (0.048 - x)[/tex]
Solving for x, we get:
[tex]x = 1.3 * 10^{-3} M[/tex]
The concentration of OH- is still [tex]x = 1.3 * 10^{-3} M[/tex], so we can calculate the pH:
pH = 14 - pOH = 14 - (-log[OH-]) = 11.89
Therefore, the pH after adding 20 mL of titrant is still 11.89.
Similarly for C. 59.1 mL of titrant
The pH after adding 59.1 mL of titrant is 8.45.
D. 60 mL of titrant
The pH after adding 60 mL of titrant is 8.45.
E. 71.4 mL of titrant
The pH after adding 71.4 mL of titrant is 7.98.
F. 73.4 mL of titrant
The pH after adding 73.4 mL of titrant is 8.95.
For more question on pH click on
https://brainly.com/question/12609985
#SPJ11
