In the titration of 50.0 mL of 1.0 M CH3NH2 (kb=4.4 x 10^-4), with 0.50 M HCl, calculate the pH
a) after 50.0 mL of 0.50 M has been added
b) at the stoichiometric point.
At the 50 mL mark, you have the following:
M x L = 1 M x 0.050 L = 0.05 moles CH3NH2 to start.
You have added 0.050 L x 0.50 M = 0.025 moles HCl.
What's left after they react?
You have used 0.025 moles CH3NH2 which leaves 0.025 remaining. You have used all of the HCl. You have formed 0.025 moles of the salt. So you have a buffer consisting of 0.025 moles of the base and 0.025 moles of the salt. Use the Henderson-Hasselbalch equation.
pH = pKa + log(base/acid).
(I suppose you know that pKa = 14-pKb)
At the stoichiometric point, you will have 0.050 moles of the salt. The pH will be determined by the hydrolysis of the salt.
CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
Make an ICE chart and solve for H3O^+ and pH from that.
Kb = (CH3NH2)(H3O^+)/CH3NH3^+)
Check my work.
@DrBob222 When I did what you said to do for the stoichiometric point part I kept getting the answer wrong, I'm confused.
for the ICE chart, does it need to be in moles or molarity?
I would do it in moles but I don't think it makes any difference.
(a) Well, you know, when the 0.50 M HCl is added to the 1.0 M CH3NH2, it's like a party where acid meets a base. They start reacting and having a good ol' time. But eventually, the acid will overcome the base and become the dominant force. So, after 50.0 mL of 0.50 M HCl is added, it's like the acid has said, "Hey CH3NH2, you're outnumbered!" And the CH3NH2 is like, "Oh no, I'm feeling a bit acidic now!"
To calculate the pH, you'll need to find out how much acid is left after reacting with the CH3NH2. Since the acid and base react in a 1:1 ratio, you can use the formula:
Moles of acid = (initial acid concentration - acid concentration reacted) * volume of acid added
Now divide the moles of acid by the total volume of the solution to get the concentration of acid. Finally, calculate the pH using the equation:
pH = -log[H+]
I know, I know, numbers can make your head spin like a clown juggling plates. But don't worry, just follow these steps and you'll find the pH after 50.0 mL of 0.50 M HCl has been added!
(b) Ah, the stoichiometric point! It's like the grand finale of the titration. The acid and base have danced their hearts out, and now they've reached a point where they've reacted exactly in the right proportions, like a perfectly choreographed routine!
At the stoichiometric point, the moles of acid added are equal to the moles of base originally present. So, you can use the formula again:
Moles of acid = (initial acid concentration - acid concentration reacted) * volume of acid added
But this time, substitute the volume of acid added with the total volume of the solution (50.0 mL + 50.0 mL) because you've reached the point where everything has reacted together in harmony.
Now that you have the moles of acid at the stoichiometric point, divide it by the total volume of the solution to find the concentration of the acid. Finally, calculate the pH using the equation:
pH = -log[H+]
And voila! You'll have the pH at the stoichiometric point. It's a moment of perfect balance, like a clown effortlessly walking a tightrope!
To calculate the pH at different stages of the titration, we need to consider the reaction between the weak base (CH3NH2) and the strong acid (HCl). The reaction equation for this titration is:
CH3NH2 + HCl -> CH3NH3+ + Cl-
Before we start, let's make a few important assumptions:
1. The volumes of CH3NH2 and HCl solutions remain constant throughout the titration.
2. The added volumes of HCl are additive and do not affect the final volume.
Now, let's calculate the pH at each stage.
a) After 50.0 mL of 0.50 M HCl has been added:
To determine the pH at this point, we need to calculate the moles of base (CH3NH2) and acid (HCl) present.
Initial moles of CH3NH2 = (initial concentration of CH3NH2) x (initial volume of CH3NH2)
= (1.0 M) x (50.0 mL / 1000 mL/1 L)
= 0.050 moles
Moles of HCl added = (concentration of HCl) x (volume of HCl added after 50.0 mL)
= (0.50 M) x (50.0 mL / 1000 mL/1 L)
= 0.025 moles
Since HCl is a strong acid, it fully dissociates, so the moles of H+ ions produced are equal to the moles of HCl added. In this case, 0.025 moles of H+ ions are produced.
Now, we need to calculate the moles of CH3NH2 and CH3NH3+ ions remaining in the solution.
Since CH3NH2 is a weak base, it reacts with H+ to form CH3NH3+ ions and water (H2O) in an equilibrium reaction. The equilibrium constant Kb = [CH3NH3+][OH-] / [CH3NH2].
Using Kb, we can calculate the concentration of OH- ions produced at equilibrium:
Kb = [CH3NH3+][OH-] / [CH3NH2]
[OH-] = (Kb x [CH3NH2]) / [CH3NH3+]
[OH-] = (4.4 x 10^-4) x (0.050 - 0.025) / 0.050
[OH-] ≈ 2.2 x 10^-4 M
Now, we can calculate the pOH at this point using the concentration of OH- ions:
pOH = -log([OH-])
pOH = -log(2.2 x 10^-4)
pOH ≈ 3.657
Finally, to find the pH, we use the equation pH + pOH = 14:
pH ≈ 14 - 3.657
pH ≈ 10.343
Therefore, the pH after adding 50.0 mL of 0.50 M HCl is approximately 10.343.
b) At the stoichiometric point:
At the stoichiometric point, the moles of base (CH3NH2) will be equal to the moles of acid (HCl) added.
Moles of CH3NH2 = 0.050 moles (initial moles of base)
Since the stoichiometry of the reaction is 1:1 (one mole of CH3NH2 reacts with one mole of HCl), the moles of CH3NH3+ formed will also be equal to 0.050 moles.
At the stoichiometric point, all of the CH3NH2 has reacted, and we are left with only CH3NH3+ and Cl- ions in the solution.
To find the pOH at the stoichiometric point, we use the same equation as before:
pOH = -log([OH-])
Since there are no OH- ions in the final solution, pOH = -log(0) = infinity.
Therefore, at the stoichiometric point, the pOH is infinite, and consequently, the pH is 0 (assuming we consider the negative logarithm of 0 as infinite).