We prepare a buffer by adding 0.5289 mol #HNO_3# to 0.5291 L of a solution containing 1.924 M trimethylamine [#(CH_3)_3N#, #Kb = 6.500x10^-5#]. What is the pH of this buffer? How do you solve this? Please help!?
Here's how you can do this:
Trimethylamine is a weak base and accepts a proton from water:
These are equilibrium concentrations.
We can use the data given to find
First we need to get the initial moles of TME:
When the nitric acid is added some of these are protonated to give the salt:
You can see that these react in a 1:1 molar ratio.
So the number of moles of TME remaining =
By the same reasoning the number of moles of the salt formed must be equal to the number of moles of nitric acid added.
Rearranging the expression for
Now we have worked out the initial moles of the base and the salt. The expression requires the no. of moles at equilibrium.
Because the value of
There may be a volume change on mixing but this does not matter since the total volume is common to both base and salt so will cancel.
This means we can put moles directly into the expression for
We know that:
Although Michael's answer is just as valid, here's another approach.
Using the Henderson-Hasselbalch equation, all we have to do is:
- Determine the
#"pKa"#, if you are using the #"pKa"#version of the equation.
- Calculate the
#-log(K_b) = "pKb" = 4.187#
Next, we consider the equation itself:
#\mathbf("pH" = "pKa" + log \frac(["B"])(["BH"^(+)]))#
#"B"#is trimethylamine and #"BH"^(+)#is protonated trimethylamine.
A nice shortcut to finding
When the added acid is strong, the
#\mathbf("mol")#s of acid added (here, it's #"HNO"_3#) is the #\mathbf("mol")#s of new conjugate acid #\mathbf("BH"^(+))#formed and the #\mathbf("mol")#s of original base #\mathbf("B")#lost.
#n_"strong acid" = n_("BH"^(+), "eq")#
#n_("B","eq") = n_("B",i) - n_"strong acid"#
Lastly, the volume of the solution is common to everything in the solution, so a molar ratio works, and you don't have to use concentrations.
Therefore, we can write the (base-10) logarithm argument as
#"pH" = "pKa" + log ((n_("B",i) - n_"strong acid")/(n_"strong acid"))#
First we have
So, our alternative method also gives us:
#color(blue)("pH") = 9.813 + log((1.018 - 0.5289)/(0.5289))#
Use whichever method you prefer.