Consider the titration of 30.0 mL sample
of 0.050 M NH3 with 0.025 M HCl.
Calculate the pH after the following volumes
of titrant (acid)have been added:
a)0 mL,
b)20.0mL,
c)60.0mL,
d)65mL
Kb for NH3 is 1.8 x 10^-5
a) After adding 0 mL of titrant, the pH will remain the same as the starting pH of the NH3 solution. So, the pH will be "No change. It's as basic as your taste in music!"
b) After adding 20.0 mL of titrant, we need to calculate the moles of NH3 and HCl reacted using the given concentrations and volumes. Then, we calculate the moles of excess HCl or NH3 left and find the concentration of each species. Finally, we use the Kb expression to calculate the pOH and convert it to pH.
c) After adding 60.0 mL of titrant, we go through the same process as in part b) to calculate the concentration of NH3 and HCl after the reaction. Then, we can calculate the pOH and convert it to pH.
d) After adding 65 mL of titrant, the solution becomes acidic due to the excess HCl. So, the pH will be determined by the concentration of HCl and can be calculated using the concentration and Kb expression.
To calculate the pH of the solution after adding different volumes of HCl, we need to consider the reaction between NH3 (ammonia) and HCl (hydrochloric acid). This is an acid-base reaction where NH3 acts as a base and HCl acts as an acid.
The balanced equation for the reaction is:
NH3 + HCl -> NH4+ + Cl-
First, let's calculate the number of moles of NH3 in the original 30.0 mL (0 mL of HCl).
a) 0 mL (original volume):
Since no HCl has been added, the concentration of NH3 remains the same. The number of moles of NH3 can be calculated using the formula: moles = concentration x volume.
moles of NH3 = 0.050 M x 0.030 L = 0.0015 moles
To calculate the initial concentration of NH4+ (NH4+ is formed when NH3 accepts a proton from HCl), we need to consider the reaction stoichiometry. It is a 1:1 ratio between NH3 and NH4+.
So, the initial concentration of NH4+ will also be 0.050 M.
Next, we need to consider the dissociation of NH4+ in water. NH4+ acts as a weak acid and donates a proton (H+) to water to form hydronium ions (H3O+).
The dissociation reaction is:
NH4+ + H2O -> H3O+ + NH3
Since NH4+ is a weak acid, we can assume it only partially dissociates, and we can use the Kb value of NH3 to calculate the concentration of H3O+.
The Kb expression for NH3 is:
Kb = [NH3][H3O+]/[NH4+]
Given that Kb for NH3 is 1.8 x 10^-5, and the initial concentration of NH4+ is 0.050 M, we can rearrange the equation to solve for [H3O+].
[H3O+] = (Kb x [NH4+]) / [NH3]
[H3O+] = (1.8 x 10^-5) x (0.050) / (0.050)
[H3O+] = 1.8 x 10^-5 M
Finally, we can calculate the pH using the formula: pH = -log[H3O+].
pH = -log(1.8 x 10^-5)
pH ≈ 4.74
So, the pH after adding 0 mL (original volume) of HCl is approximately 4.74.
To calculate the pH after adding 20.0 mL of HCl (b):
Since 20.0 mL of HCl has been added, the volume of the solution is now 30.0 mL + 20.0 mL = 50.0 mL.
To find the moles of HCl added, we can use the formula: moles = concentration x volume.
moles of HCl = 0.025 M x 0.020 L = 0.0005 moles
Since HCl is a strong acid, it completely dissociates in water, providing an equal number of H+ ions.
Therefore, after the addition of HCl, the number of H3O+ ions will be 0.0005 moles.
The total volume of the solution is now 50.0 mL, so the new concentration of H3O+ is:
[H3O+] = (0.0005 moles) / (0.050 L)
[H3O+] = 0.01 M
pH = -log(0.01)
pH = 2
So, the pH after adding 20.0 mL of HCl is 2.
You can follow the same procedure to calculate the pH for the other volumes of HCl added (c and d) using their respective volumes.
The secret to these is to recognize what you have in the solution at each addition.
The first thing you do is to determine where the equivalence point is; i.e., number of mL HCl added to get to the equivalence point.
20 mL--you have a mixture of NH3 and NH4Cl. That makes a buffer; use the Henderson-Hasselbalch equation.
60 mL is the eq point. The pH is determined by the hydrolysis of the salt NH4Cl.
0 mL you have 0.50M NH3. I'm sure you've worked many a problem like that.