# How would you use the Hendersonâ€“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

##### 1 Answer

#### Answer:

Here's what I would do.

#### Explanation:

You're dealing with a **buffer solution** that contains *pyridine*, **weak base**, and *pyridinium chloride*, **conjugate acid**, the pyridinium cation,

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 = 1.7 * 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_"solution" ~~ rho_"water" ~~ "1 g mL"^(-1)#

Now, let's assume that you're dealing with a

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^3"mL")color(white)(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 **by mass** pyridine and **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 **mole ratio** to form pyridinium cations and chloride anions

#"C"_ 5"H"_ 5"NHCl"_ ((aq)) -> "C"_ 5"H"_ 5"NH"_ ((aq))^(+) + "Cl"_((aq))^(-)#

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

#["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")))))#

#"pOH" = 8.77 + (-0.0721) = 8.70#

Since you know that at room temperature you have

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#

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

#"pH" = 14 - 8.70 = color(green)(|bar(ul(color(white)(a/a)5.30color(white)(a/a)|)))#