In the titration of 77.5 mL of 1.0 M methylamine, CH3NH2 (Kb = 4.4 10-4), with 0.38 M HCl, calculate the pH under the following conditions.
(a) after 50.0 mL of 0.38 M HCl has been added
(b) at the stoichiometric point
b. pH is due to the hydrolysis of the salt. You must first determine the M of the salt. That will be moles base x M base divided by the total volume (base+ acid) in liters. I will call that value x.
........CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
initial...x..................0........0
change....-y.................y........y
equil....x-y..................y.......y
Ka = (Kw/Kb for CH3NH2) = (H3O^+)(CH3NH2)/(CH3NH3^+)
(1E-14/Kb) = (y)(y)/(x)
Solve for x = (H3O^+) and convert to pH.
a.
77.5 mL x 1.0M = M x L = 0.0775 mols CH3NH2.
50 mL x 0.38 = = 0.019 moles.
.........CH3NH2 + HCl ==>CH3NH3Cl+ H2O
initial...0.0775...0.......0
added...........0.019............
change.0.0775-0.019 -0.019.+0.019
equil...-0.0585...0........0.019
So you have a solution that is
(CH3NH2) = 0.0595/total volume
(CH3NH3Cl) = 0.019/total volume
This is a buffered solution. Plug these values into the buffer equation (the Henderson-Hasselbalch equation) or into the Kb expression and solve for pH.
Thank You
To calculate the pH under the given conditions, we need to consider the reaction between methylamine (CH3NH2) and HCl.
The balanced chemical equation for the reaction is:
CH3NH2 + HCl → CH3NH3+ + Cl-
(a) After 50.0 mL of 0.38 M HCl has been added:
Step 1: Calculate the moles of HCl added:
Moles of HCl = volume (L) x concentration (M)
Moles of HCl = 0.050 L x 0.38 M = 0.019 moles
Step 2: Calculate the initial moles of CH3NH2:
Moles of CH3NH2 = volume (L) x concentration (M)
Moles of CH3NH2 = 0.0775 L x 1.0 M = 0.0775 moles
Step 3: Determine the limiting reagent:
Since the mole ratio between HCl and CH3NH2 is 1:1, the limiting reagent is CH3NH2.
Step 4: Calculate the remaining moles of CH3NH2:
Remaining moles of CH3NH2 = Initial moles - Moles of HCl reacted
Remaining moles of CH3NH2 = 0.0775 moles - 0.019 moles = 0.0585 moles
Step 5: Calculate the concentration of CH3NH3+:
Concentration of CH3NH3+ = Remaining moles / Volume (L)
Concentration of CH3NH3+ = 0.0585 moles / (0.0775 L + 0.050 L) = 0.425 M
Step 6: Calculate the pOH:
pOH = -log10 (OH- concentration)
Since this is a weak base reaction, the concentration of OH- can be calculated using Kb:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Solving for [OH-], we get:
[OH-] = (Kb x [CH3NH2]) / [CH3NH3+]
[OH-] = (4.4 x 10^-4) x (0.0585 M) / (0.425 M)
[OH-] = 6.044 x 10^-5 M
pOH = -log10 (6.044 x 10^-5) = 4.218
Step 7: Calculate the pH:
pH = 14 - pOH
pH = 14 - 4.218 = 9.782
Therefore, the pH after adding 50.0 mL of 0.38 M HCl is approximately 9.782.
(b) At the stoichiometric point:
At the stoichiometric point, all the CH3NH2 is completely consumed and converted into CH3NH3+.
Step 1: Calculate the moles of HCl required to reach the stoichiometric point:
Moles of HCl = volume (L) x concentration (M)
Moles of HCl = 0.050 L x 0.38 M = 0.019 moles
Step 2: Calculate the moles of CH3NH2 reacted at the stoichiometric point:
Moles of CH3NH2 reacted = Moles of HCl required
Step 3: Calculate the remaining moles of CH3NH2:
Remaining moles of CH3NH2 = Initial moles - Moles of CH3NH2 reacted
Remaining moles of CH3NH2 = 0.0775 moles - 0.019 moles = 0.0585 moles
Step 4: Calculate the concentration of CH3NH3+:
Concentration of CH3NH3+ = Remaining moles / Volume (L)
Concentration of CH3NH3+ = 0.0585 moles / (0.0775 L + 0.050 L) = 0.425 M
Step 5: Calculate the pOH:
pOH = -log10 (OH- concentration)
Using the equation from step (a), we already know the [OH-] concentration is 6.044 x 10^-5 M.
pOH = -log10 (6.044 x 10^-5) = 4.218
Step 6: Calculate the pH:
pH = 14 - pOH
pH = 14 - 4.218 = 9.782
Therefore, the pH at the stoichiometric point is approximately 9.782.
To calculate the pH at each condition, we need to consider the reaction between methylamine (CH3NH2) and HCl. Methylamine is a weak base, and HCl is a strong acid. The reaction between them can be represented as:
CH3NH2 + HCl -> CH3NH3+ + Cl-
(a) The equation shows that one mole of methylamine reacts with one mole of HCl to form CH3NH3+. Therefore, to calculate the pH after adding 50.0 mL of 0.38 M HCl, we need to find out how much of the methylamine has reacted.
First, calculate the moles of HCl used:
Moles of HCl = volume (L) x concentration (mol/L)
Moles of HCl = 0.050 L x 0.38 mol/L
Next, we need to consider the moles of methylamine that have reacted. Since the mole ratio between methylamine and HCl is 1:1, the number of moles of methylamine that have reacted is equal to the moles of HCl used.
Now, we can calculate the concentration of methylamine remaining in the solution:
Initial moles of methylamine = volume (L) x concentration (mol/L)
Initial moles of methylamine = 0.0775 L x 1.0 mol/L
Moles of methylamine remaining = Initial moles of methylamine - Moles of HCl used
Finally, we can calculate the concentration of methylamine (CH3NH3+) in the solution:
Concentration of methylamine (CH3NH3+) = moles of methylamine remaining / volume (L)
Using the concentration of CH3NH3+, we can calculate the pOH and then the pH of the solution.
(b) The stoichiometric point is the point at which the reactants are in stoichiometric proportions, meaning that all the methylamine has reacted with HCl. Therefore, at the stoichiometric point, there is no more methylamine left in the solution. We can assume that all the methylamine has been converted to CH3NH3+, and the concentration of CH3NH3+ at the stoichiometric point is equal to the initial concentration of methylamine (1.0 M).
Again, using the concentration of CH3NH3+, we can calculate the pOH and then the pH of the solution at the stoichiometric point.