Question

Consider the titration of a 28.0?mL sample of 0.180M CH3NH2 with 0.150M HBr. Determine each of the following.

A.) the initial pH

B.) the volume of added acid required to reach the equivalence point

C.) the pH at 6.0mL of added acid

D.) the pH at one-half of the equivalence point

E.) the pH at the equivalence point

F.) the pH after adding 6.0mL of acid beyond the equivalence point

Some help & direction would be great very confused! Thank you!

Answer 1

This problem should be worth way more than 10 points, but here goes:

The reaction between CH3NH2 and HBr: . . .CH3NH2 + HBr ==> CH3NH3+ + Br-?

Once you start to add HBr to the CH3NH2, you make CH3NH3+, which is the conjugate acid of CH3NH2. You now have a buffer system (a weak base and its conjugate acid). The pH up until the endpoint will be determined by the Henderson-Hasselbalch equation for buffers.

PART A.

We know that: pH = -log [H+]

So we need to find [H+] to get the pH of the solution. But initially, we only have [OH-] in solution.

However, because of the autoionization of water, we know that:
[H+][OH-] = Kw = 1 x 10^-14
So now we have our link between [H+] and [OH-].

Let's plug in what we know.

Initial [OH-] = 0.180 M OH- , therefore:

[H+] (0.180M) = 1x10^-14

which gives us [H+] = 5.56 x 10^-14 M
(Because Kw is a constant, I'm sticking with the 3 sig figs that 0.180 M gives us for our answer) This value should make sense because with such a high [OH-], there should be very little [H+].

So pH = -log (5.56 x 10^-14M) = 13.255

PART B.

at the equivalent point in a titration, moles of acid = moles of base.

moles (n) = C x V (C is concentration and V is volume in Liters).

n= 0.180 M OH- x 0.0280 L = 0.00504 moles OH- initially present.

So at equilibrium, to neutralize 0.00504 moles base requires 0.00504 moles acid added.

To find the Volume of acid, use the moles equation again using the concentration of acid used:

0.00504 moles H+ = (0.150M H+) x V, so V=0.0336L or 33.6 mL

PART C

In order to reach the equivalence point, we just found out we need to add 33.6 mL of the 0.150M HCl. Here, we've only added 6 mL - so we have not reached the equivalence point yet - but we have neutralized SOME of the moles of base initially present.

lets first find the # of moles of acid we are adding here:

n = C x V; n = (0.150M H+)(0.005L) ===> 0.00075 moles H+ added.

Remember our initial moles of base was 0.00504 moles. By adding 0.0075 moles of acid, we neutralize the same moles of base. To find what is left over subtract what we added from what was initially present:

0.00504 moles - 0.00075 moles = 0.00429 moles OH- left.

In order to find our new concentration, we need the total volume now:

0.0280L + 0.006L = 0.022L

C = n/V; [OH-] = 0.00429 moles OH-/ 0.022L

[OH-] = 0.195 M and substituting this into our pH equation PROPERLY:

pH = -log ( 1x10^-14 / 0.0195 M) = 12.29

This is the pH after adding 6 mL of our 0.150 M HCl.

PART D.

At the equivalence point: moles base = moles of acid. we have no moles of either left over , but because of the autoionization of water, [H+] = 1 x10^-7, so pH = 7.

PART E.

To find the pH after the equivalence point, we need the [H+] after the equivalence point, so we need to know how many moles of acid are present and our total volume of solution to find this concentration value.

moles = (0.150 M HCl) x (0.006L) = 0.0009 moles H+

our volume started at 28.0mL, then we added 33.6 mL to get to the eq. point (part b above). Now we are adding 6 mL beyond that - giving us a total of 67.6 mL or 0.0676L.

our [H+] then = (0.0009 moles) / (0.0676L) = 0.0133 M H+

Then:

pH = -log (0.013 M H+) = 1.88

Our pH is now 1.88.