# How would you use the Henderson–Hasselbalch equation to calculate the pH of the solution? A solution that is 0.800 % C_5H_5N by mass, and 0.990 % C_5H_5NHCl by mass

May 17, 2016

Here's what I would do.

#### Explanation:

You're dealing with a buffer solution that contains pyridine, $\text{C"_5"H"_5"N}$, a weak base, and pyridinium chloride, $\text{C"_5"H"_5"NHCl}$, the salt of its conjugate acid, the pyridinium cation, ${\text{C"_5"H"_5"NH}}^{+}$.

In order to be able to use the Henderson - Hasselbalch equation, which for a buffer that contains a weak base and its conjugate looks like this

color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))

you will need to determine the concentrations of pyridine and of the pyridium cation. The value of the base dissociation constant, ${K}_{b}$, for pyridine can be found here

${K}_{b} = 1.7 \cdot {10}^{- 9}$

http://www.bpc.edu/mathscience/chemistry/table_of_weak_bases.html

Now, the problem doesn't provide you with the density of the solution; however, because you're dealing with such small amounts of pyridine and pyridinium chloride, you can assume that the density of the solution is approximately equal to that of water.

To keep the calculations simple, you should take

${\rho}_{\text{solution" ~~ rho_"water" ~~ "1 g mL}}^{- 1}$

Now, let's assume that you're dealing with a $\text{1-L}$ sample of this buffer solution. Since you know that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 L" = 10^3"mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you can say that this sample will be equivalent to

1 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = 10^3"mL"

Now, this sample will have a mass of

10^3color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = 10^3"g"

The solution is said to be 0.800% by mass pyridine and 0.990% by mass pyridinium chloride. Use these concentrations to find the mass of the two chemical species in this sample

10^3color(red)(cancel(color(black)("g solution"))) * ("0.800 g C"_5"H"_5"N")/(100color(red)(cancel(color(black)("g solution")))) = "8.00 g C"_5"H"_5"N"

10^3color(red)(cancel(color(black)("g solution"))) * ("0.990 g C"_5"H"_5"NHCl")/(100color(red)(cancel(color(black)("g solution")))) = "9.90 g C"_5"H"_5"NHCl"

Use the molar masses of the two compounds to determine how many moles of each you have present

8.00 color(red)(cancel(color(black)("g"))) * ("1 mole C"_5"H"_5"N")/(79.1color(red)(cancel(color(black)("g")))) = "0.10114 moles C"_5"H"_5"N"

9.90color(red)(cancel(color(black)("g"))) * ("1 mole C"_5"H"_5"NHCl")/(115.56color(red)(cancel(color(black)("g")))) = "0.085670 moles C"_5"H"_5"NHCl"

Now, pyridinium chloride dissociates in a $1 : 1$ mole ratio to form pyridinium cations and chloride anions

${\text{C"_ 5"H"_ 5"NHCl"_ ((aq)) -> "C"_ 5"H"_ 5"NH"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

This means that for every mole of pyridinium chloride that you dissolve in solution, you get one mole of pyridinium cations.

Since this sample has a total volume of $\text{1 L}$, the molarity of the two species will be

["C"_5"H"_5"N"] = "0.10114 moles"/"1 L" = "0.10114 M"

["C"_5"H"_5"NH"^(+)] = "0.085670 moles"/"1 L" = "0.085670 M"

Use the Henderson - Hasselbalch equation to find the pOH of the buffer

"pOH" = - log(K_b) + log( (["C"_5"H"_5"NH"^(+)])/(["C"_5"H"_5"N"]))

Plug in your values to find

"pOH" = -log(1.7 * 10^(-9)) + log( (0.085670 color(red)(cancel(color(black)("M"))))/(0.10114color(red)(cancel(color(black)("M")))))

$\text{pOH} = 8.77 + \left(- 0.0721\right) = 8.70$

Since you know that at room temperature you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you can say that the pH of the solution will be equal to

$\text{pH} = 14 - 8.70 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 5.30 \textcolor{w h i t e}{\frac{a}{a}} |}}}$