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.**

Consider the titration of a 26.0-mL sample of 0.170 M CH3NH2
with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)
Initial pH- 11.94
Volume of added acid required to reach the equivalence point:
30.5 mL
Determine the pH after adding 5.0 mL of acid beyond the
equivalence point.

Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2
with 0.155 M HBr. Determine each of the following.
a. the pH at one-half of the equivalence point
b. the pH at the equivalence point
c. the pH after adding 6.0 mL of acid beyond the equivalence
point
i already found that the initial pH is 11.95, the volume of acid
added to reach equivelance point is 29.0mL, and the pH og 6.0mL of
added acid is 11.23

Consider the titration of a 26.0-mL sample of 0.180 M CH3NH2
with 0.145 M HBr. (The value of Kb for CH3NH2 is
4.4×10−4.)
Part A
Determine the initial pH.
Part B
Determine the volume of added acid required to reach the
equivalence point. answer to 3 sig figs
Part C
Determine the pH at 6.0 mL of added acid
Part D
Determine the pH at one-half of the equivalence point.
Part E
Determine the pH at the equivalence point.
Part...

Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2
with 0.150 M HBr. Determine each of the following.
1. the pH at 4.0 mL of added
acid
2.the pH at the equivalence point
3.the pH after adding 6.0 mL of acid beyond the equivalence
point

Consider the titration of a 27.0 −mL sample of 0.175 MCH3NH2
with 0.150 M HBr. Express answers using two decimal places.
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 6.0 mL 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 4.0 mL of acid beyond the equivalence
point

Consider the titration of a 27.0 −mL sample of 0.180 M CH3NH2
with 0.145 M HBr. Determine the pH after adding 6.0 mL of acid
beyond the equivalence point.

Consider the titration of a 81.9 mL sample of 0.172 M CH3NH2 ,
methylamine, with 0.2 M HBr. The Kb of methylamine is 4.4x10−4.
Determine the volume (mL) of added acid required to reach the
equivalence point

Consider the titration of a 25.0?mL sample of 0.175M CH3NH2 with
0.145M HBr. Determine each of the following.
the pH at 6.0mL of added acid

Consider the titration of a 27.8 −mL sample of 0.125 M RbOH with
0.110 M HCl. 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 6.0 mL of added acid D) the pH at the equivalence
point E) the pH after adding 4.0 mL of acid beyond the equivalence
point

Consider the titration of a 23.4 −mL sample of 0.125 M RbOH with
0.100 M HCl. 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 5.3 mL of added acid
D) the pH at the equivalence point
E) the pH after adding 5.4 mL of acid beyond the equivalence
point

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