# 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!?

##### 2 Answers

Here's how you can do this:

#### Explanation:

Trimethylamine is a weak base and accepts a proton from water:

For which:

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. - Find
#["BH"^(+)]# and#["B"]# . - Calculate the
#"pH"# from there.

First:

#-log(K_b) = "pKb" = 4.187#

which means

Next, we consider the equation itself:

#\mathbf("pH" = "pKa" + log \frac(["B"])(["BH"^(+)]))# where

#"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 ofnewconjugate acid#\mathbf("BH"^(+))# formedand the#\mathbf("mol")# s oforiginalbase#\mathbf("B")# lost.That is,

#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))#

#~~ color(blue)(9.78)#

Use whichever method you prefer.