I have another similar problem. I have calculated everything given (Correctly), but I can't seem to get the right pH for the stoichiometric point (I keep getting 7.16)
Question:
Suppose that 50.0 mL of 0.25 M CH3NH2(aq) is titrated with 0.35 M HCl(aq). Refer to table 1 and table 2.
(e) What volume of 0.35 M HCl(aq) is required to reach the stoichiometric point?
35.7 mL (correct answer)
(f) Calculate the pH at the stoichiometric point.
I set up the "PUG" Chart, found my concentrations of the base to be 5.83E-5M and the acid .146M
I then used the expression:
HB^+ +H2O <--> H3O^+ + B to set up the ICE chart. I then ended up with
Ka= (1E-14)/(3.6E-4)=(5.83E-5)x / (.1458)
since I found the change in x to be negligble. 3.6E-4 is the given Kb for the base.
Could anyone shed some light on my calculations?
Thanks!
I think your problem is that you have substituted 5.83E-5 for the base while letting x stand for (H3O^+). x is ALSO (CH3NH2) at that point. The base, at that point, is ONLY the amount from the hydrolysis of the salt. I agree that x in the denominator is small and can be neglected. I think the pH at the equivalence point is 5.69 or close to that. 0.1458 is concn of the salt. By the way, what is a PUG chart. I find most student know what an ICE chart is; perhaps they know a PUG chart, too, but I don't.
Based on the information you provided, it seems that you have set up the correct ice chart and used the correct expression for the equilibrium constant (Ka). However, there might be a slight error in your calculation.
Let's break down the steps to calculate the pH at the stoichiometric point:
1. Determine the concentration of the base (CH3NH2) and the acid (HCl) at the stoichiometric point:
From your calculations, the concentration of the base (CH3NH2) at the stoichiometric point is 5.83E-5 M. The concentration of the acid (HCl) at the stoichiometric point can be calculated using the volume and molarity provided in the question. Since 35.7 mL of 0.35 M HCl(aq) is required to reach the stoichiometric point, we can use the dilution formula:
C1V1 = C2V2
Where C1 is the initial concentration of the acid, V1 is the initial volume of the acid, C2 is the final concentration of the acid, and V2 is the final volume (the sum of the base and the acid volumes).
Plugging in the values: (0.35 M)(V1) = (0.35 M)(35.7 mL + 50.0 mL)
Solve for V1: V1 = [(0.35 M)(35.7 mL + 50.0 mL)] / (0.35 M)
V1 = 50.0 mL
So, the concentration of the acid at the stoichiometric point is also 0.35 M.
2. Use the equilibrium expression to determine the concentration of H3O+ at the stoichiometric point:
Since CH3NH2 is a weak base, it reacts with water to produce CH3NH3+ (the conjugate acid) and OH- (hydroxide ions). At the stoichiometric point, the concentration of CH3NH3+ is equal to the concentration of OH-.
Therefore, the concentration of OH- at the stoichiometric point is also 5.83E-5 M.
We can use this concentration of OH- to calculate the concentration of H3O+ using the Kw, the equilibrium constant for water:
Kw = [H3O+][OH-] = 1.0 x 10^-14
[H3O+] = Kw / [OH-] = (1.0 x 10^-14) / (5.83E-5)
[H3O+] = 1.716 x 10^-10 M
3. Calculate the pH using the concentration of H3O+:
pH = -log[H3O+]
pH = -log(1.716 x 10^-10)
pH = 9.766
Therefore, the pH at the stoichiometric point should be approximately 9.766, which is different from the value you obtained (7.16).