A 40.00 mL sample of 0.1 M NH3 is titrated with 0.150 M HCl solution. Kb = 1.8 * 10^-5
What volume of base is required to reach the equivalence point?
What is the ammonium ion concentration at the equivalence point?
What is the pH of the solution at the equivalence point?
How much of this do you know how to do and exactly what is it you don't understand.?
I think that the first question you just do MaVa = MbVb (a is acid b is base). I have no idea how to do the second question, I'm guessing that it's just .1 M. I also know that pH = -log [H+], but do I use .15M? I'm just confused with the equivalence point.
right. That's how you do a.
Thre second part. You know how many mols it takes to neutralize the ammonia. That will be the mols of NH4Cl produced. And that divided by the volume (in liters which will be the sum of volume HCl + volume NH3) will be the molarity of the ammonium ion concn at the equivalence point (which is the neutralization point).
For the last part, you know the concn of the ammonium ion. Just hydrolye that (react with water) and do the ICE thing.
NH4^+ + HOH ==> NH3 + H3O^+
Ka = Kw/Kb = (H3O^+)(NH3)/(NH4^+)
What is the ICE thing? Is it pH = Ka + log [HA]/[A-]?
The ICE thing is difficult to show on these boards because we have a problem with spacing. But I think I can show you.
I re-read the problem and I think it means something different. I think it should have said, "what volume of ACID (not base) is required to reach the equivalence point?"
For a) you should have 26.67 mL HCl required to neutralize 40 mL of 0.1 M NH3 solution.
b)The ammonium ion concn at the equivalence point is 0.004 mols NH4^+/0.0667 L = 0.06 M
c)The NH4^+ hydrolyzes in water.
NH4^+ + H2O ==> NH3 + H3O^+
For the ICE we will have three rows
I
C
E
but then I can't space to put what these values are under the appropriate symbols. So what I will do is write out the formula and the concentration like this.
Initial concn (before rxn with H2O):
(NH4^+) = 0.06 M
(H3O^+) = 0
(NH3) = 0
Change in concn:
(H3O^+) = x
(NH3) = x
(NH4^+) = - x
Equilibrium:
(H3O^+) = 0 + x = x
(NH3) = 0 + x = x
(NH4^+) = 0.06 - x = 0.06 - x
Then Ka = Kw/Kb = (H3O^+)(NH3)/(NH4^+)
You know Kw and Kb. Substitute x for (H3O^+) and for (NH3) and substitute 0.06 - x for (NH4^+). solve for x which equals (H3O^+), then pH = - log (H3O^+),
You need not be intimidated with the word equivalence point. That is just the point at which the acid and base have exactly neutralized each other. In this case we have the reaction:
NH3 + HCl ==> NH4Cl
So we have no NH3 remaining, we have no HCl remaining (because both have been neutralized) and we have a solution of NH4Cl in water.
To determine the volume of base required to reach the equivalence point in this titration, we need to use the concept of stoichiometry. The balanced chemical equation for the reaction between NH3 (base) and HCl (acid) is:
NH3 + HCl → NH4Cl
From the equation, we can see that one mole of NH3 reacts with one mole of HCl to form one mole of NH4Cl. Therefore, the number of moles of HCl required to react completely with NH3 can be calculated as:
Moles of HCl = (0.1 M) x (0.040 L) = 0.004 mol
Since the HCl concentration is given as 0.150 M, we can use the following relationship to find the volume of HCl required:
Volume of HCl (L) = Moles of HCl / HCl concentration
Volume of HCl (L) = 0.004 mol / 0.150 M = 0.0267 L = 26.7 mL
So, the volume of HCl required to reach the equivalence point is 26.7 mL.
At the equivalence point, all the moles of NH3 have reacted completely with HCl. Therefore, the concentration of NH4+ ions (ammonium ion) will be equal to the initial concentration of NH3. In this case, the initial concentration of NH3 is 0.1 M.
Thus, at the equivalence point, the ammonium ion concentration is 0.1 M.
To determine the pH of the solution at the equivalence point, we need to consider the reaction between NH3 and HCl. When NH3 reacts with HCl, NH4+ ions are formed, which are acidic in nature. Therefore, we need to calculate the concentration of NH4+ ions to determine the pH.
Since we already know that the concentration of NH4+ ions at the equivalence point is 0.1 M, we can use the equilibrium expression for the ammonium ion dissociation reaction:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant for this reaction Kb is given as 1.8 x 10^-5.
Using the expression for Kb, we can calculate the concentration of H3O+ ions at the equivalence point. Let's denote the concentration of H3O+ ions as [H3O+].
Kb = [NH3][H3O+] / [NH4+]
1.8 x 10^-5 = (0.1)[H3O+] / (0.1)
[H3O+] = 1.8 x 10^-5 M
Since [H3O+] is equal to [H+], the pH of the solution at the equivalence point is given by:
pH = -log[H+]
pH = -log(1.8 x 10^-5) ≈ 4.74
Therefore, the pH of the solution at the equivalence point is approximately 4.74.