# 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!?

Aug 4, 2016

$\textsf{p H = 9.78}$

Here's how you can do this:

#### Explanation:

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

$\textsf{{\left(C {H}_{3}\right)}_{3} N + {H}_{2} O r i g h t \le f t h a r p \infty n s {\left(C {H}_{3}\right)}_{3} N {H}^{+} + O {H}^{-}}$

For which:

$\textsf{{K}_{b} = \frac{\left[{\left(C {H}_{3}\right)}_{3} N {H}^{+}\right] \left[O {H}^{-}\right]}{\left[{\left(C {H}_{3}\right)}_{3} N\right]} = 6.500 \times {10}^{- 3} \textcolor{w h i t e}{x} \text{mol/l}}$

These are equilibrium concentrations.

We can use the data given to find $\textsf{\left[O {H}^{-}\right]}$ and then use the ionic product of water to find $\textsf{\left[{H}^{+}\right]}$ hence the pH.

First we need to get the initial moles of TME:

$\textsf{{n}_{T M E} = c \times v = 1.924 \times 0.5291 = 1.018}$

When the nitric acid is added some of these are protonated to give the salt:

$\textsf{{\left(C {H}_{3}\right)}_{3} N + {H}^{+} \rightarrow {\left(C {H}_{3}\right)}_{3} N {H}^{+}}$

You can see that these react in a 1:1 molar ratio.

So the number of moles of TME remaining = $\textsf{1.018 - 0.5289 = 0.4890}$

By the same reasoning the number of moles of the salt formed must be equal to the number of moles of nitric acid added.

$\therefore$ no. moles of $\textsf{{\left(C {H}_{3}\right)}_{3} N {H}^{+}}$ formed $\textsf{= 0.5289}$

Rearranging the expression for $\textsf{{K}_{b} \Rightarrow}$

$\textsf{\left[O {H}^{-}\right] = {K}_{b} \times \frac{\left[{\left(C {H}_{3}\right)}_{3} N\right]}{\left[{\left(C {H}_{3}\right)}_{3} N {H}^{+}\right]}}$

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 $\textsf{{K}_{b}}$ is small I am going to assume that the initial moles of base and salt are a good approximation for the equilibrium moles.

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 $\textsf{\left[O {H}^{-}\right] \Rightarrow}$

$\textsf{\left[O {H}^{-}\right] = 6.500 \times {10}^{- 5} \times \frac{0.4890}{0.5289} = 6.010 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p O H = - \log \left[O {H}^{-}\right] = - \log \left[6.010 \times {10}^{- 5}\right] = 4.221}$

We know that:

$\textsf{p H + p O H = 14}$ at $\textsf{{25}^{\circ} \text{C}}$

$\therefore$$\textsf{p H = 14 - p O H = 14 - 4.221 = 9.78}$

Aug 4, 2016

Although Michael's answer is just as valid, here's another approach.

Using the Henderson-Hasselbalch equation, all we have to do is:

1. Determine the $\text{pKa}$, if you are using the $\text{pKa}$ version of the equation.
2. Find $\left[{\text{BH}}^{+}\right]$ and $\left[\text{B}\right]$.
3. Calculate the $\text{pH}$ from there.

First:

$- \log \left({K}_{b}\right) = \text{pKb} = 4.187$

which means color(green)("pKa") = "pKw" - "pKb" = 14 - 4.187 = color(green)(9.813).

Next, we consider the equation itself:

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

where $\text{B}$ is trimethylamine and ${\text{BH}}^{+}$ is protonated trimethylamine.

A nice shortcut to finding $\left[\text{B}\right]$ and $\left[{\text{BH}}^{+}\right]$ is...

When the added acid is strong, the $\setminus m a t h b f \left(\text{mol}\right)$s of acid added (here, it's ${\text{HNO}}_{3}$) is the $\setminus m a t h b f \left(\text{mol}\right)$s of new conjugate acid $\setminus m a t h b f \left({\text{BH}}^{+}\right)$ formed and the $\setminus m a t h b f \left(\text{mol}\right)$s of original base $\setminus m a t h b f \left(\text{B}\right)$ 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 color(green)(\frac(["B"])(["BH"^(+)]) = (n_("B","eq"))/(n_("BH"^(+),"eq")) = (n_("B",i) - n_"strong acid")/(n_"strong acid")):

"pH" = "pKa" + log ((n_("B",i) - n_"strong acid")/(n_"strong acid"))

First we have ${n}_{\text{B",i) = "0.5291 L" xx "1.924 mols/L" = color(green)("1.018 mols}}$. Then we were given n_"strong acid" = color(green)("0.5289 mols") monoprotic strong acid, ${\text{HNO}}_{3}$, which adds $1 : 1$.

So, our alternative method also gives us:

$\textcolor{b l u e}{\text{pH}} = 9.813 + \log \left(\frac{1.018 - 0.5289}{0.5289}\right)$

$\approx \textcolor{b l u e}{9.78}$

Use whichever method you prefer.