Consider the titration of a 10 mL aliquot of 0.1 M NH3 (pKb = 4.78) with a 0.1 M HCl.Calculate the expected pH at the different points in the titration as indicated below. Youshould show all your workinga) the start of the titration
Question
Consider the titration of a 10 mL aliquot of 0.1 M NH3 (pKb = 4.78) with a 0.1 M HCl.Calculate the expected pH at the different points in the titration as indicated below. Youshould show all your workinga) the start of the titration
Solution
a) At the start of the titration, no HCl has been added yet, so we only have NH3 in the solution. NH3 is a weak base, so it will react with water to form NH4+ and OH-. However, because it is a weak base, only a small amount of it will react.
The reaction is as follows:
NH3 + H2O ⇌ NH4+ + OH-
We can use the Kb expression to find the concentration of OH- in the solution:
Kb = [NH4+][OH-] / [NH3]
We know that Kb = 10^-pKb = 10^-4.78 = 1.66 x 10^-5
At the start of the titration, [NH3] is 0.1 M, and because the reaction hasn't proceeded far, [NH4+] and [OH-] are approximately 0.
Setting up the Kb expression gives us:
1.66 x 10^-5 = x^2 / 0.1
Solving for x gives us the concentration of OH- in the solution:
x = sqrt(1.66 x 10^-5 * 0.1) = 0.00129 M
We can then use the OH- concentration to find the pOH of the solution:
pOH = -log[OH-] = -log(0.00129) = 2.89
Finally, we can find the pH of the solution using the relationship pH + pOH = 14:
pH = 14 - pOH = 14 - 2.89 = 11.11
So, at the start of the titration, the pH of the solution is approximately 11.11.
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