Equilibrium and Equivalence Calculation [Very Difficult]

[iBOOM]

Homework Statement

1. What volume of .022M HCl must be added to the 110ml buffer solution to completely react with the ammonia (NH₃) to reach equivalence. Then, what is the pH of this at the equivalence point.

Buffer solution (equilibrium): NH₃ + H₂0 ⇌ NH₄⁺ + OH^-
Other Info: .045M NH₄⁺ and .024M NH₃
and the pH of the buffer is 8.99
K_{b NH3} = 1.8 x 10^-5 Not sure if this is needed though.

Homework Equations

Molarity = n(moles)/V(volume in L)
pH = -log[H⁺]
Brackets means the concentration of the substance on the inside.
Not sure what else, I am sure there's some stoichiometry involved.

The Attempt at a Solution

I'm having a lot of trouble with this one.

 

[iBOOM]

So I've actually managed to do a bit but I'd really appreciate some help.

I've got the hydrolysis equation:
NH₃ + H⁺ <--> NH₄⁺
__.024M______.022M_______________.045M
__110ml________?ml________________110ml
.00264mol_________________________.00495mol

Do i just use stoichiometry to figure out the missing volume? And the equivalence, how does that play into the problem? Thanks in advance.

 

[iBOOM]

So I ended up doing some stoichiometry and got 120 ml of HCl required and then calculated the pH to 1.96. Can someone help me out?

 

[Borek]

This is a titration problem. Equivalence point is the moment when you should end the titration. You have a solution that contains ammonia (and some NH₄⁺ which we can for now safely ignore). How much HCl is needed to reach the equivalence point? This is a simple stoichiometry.

Next part is just a question about the equivalence point pH.

 

[DeeNos]

I know that at the equivalence point, the moles of acid added will be equal to the moles of base present in the buffer solution. However, I'm not sure how to use this information to calculate the volume of acid needed.

First, we can start by writing out the balanced chemical equation for the reaction between HCl and NH3:
HCl + NH3 → NH4+ + Cl-

Next, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution at the given concentrations:
pH = pKa + log([A-]/[HA])
where pKa is the negative log of the acid dissociation constant (Ka) and [A-] and [HA] are the concentrations of the conjugate base and acid, respectively.

Since the buffer is made up of NH3 and NH4+, we can use the Kb value given to calculate the pKa:
pKa = 14 - pKb = 14 - (-log(1.8 x 10^-5)) = 14 + 5.74 = 19.74

Now, we can plug in the given concentrations to the Henderson-Hasselbalch equation:
8.99 = 19.74 + log([NH4+]/[NH3])
Solving for [NH4+]/[NH3], we get a ratio of 0.045/0.024 = 1.875.

Since we know that at the equivalence point, the moles of acid added will be equal to the moles of base present in the buffer solution, we can set up an equation using the molarity formula:
Molarity of acid = Molarity of base
nHCl/VHCl = nNH3/VNH3
where n is the number of moles and V is the volume in liters.

We can rearrange the equation to solve for VHCl:
VHCl = (nNH3/nHCl) * VNH3
Since we want to reach equivalence, we know that the moles of acid added (nHCl) will be equal to the moles of NH3 present in the buffer (nNH3). So, we can simplify the equation to:
VHCl = VNH3

Now, we can plug in the given values for the volume of the buffer (110 ml = 0.11 L) and the concentration of NH3 (0.024 M) to solve