# How would you use the Hendersonâ€“Hasselbalch equation to calculate the pH of each a solution that contains 1.40% C2H5NH2 by mass and 1.18% C2H5NH3Br by mass?

Nov 17, 2015

$\text{pH} = 11.27$

#### Explanation:

You're dealing with a buffer solution that contains ethylamine, ${\text{C"_2"H"_5"NH}}_{2}$, a weak base, and ethylammonium bromide, $\text{C"_2"H"_5"NH"_3"Br}$, the salt of its conjugate acid, the ethylammonium ion, ${\text{C"_2"H"_5"NH}}_{3}^{+}$.

The Henderson - Hasselbalch equation for a weak base - conjugate acid buffer looks like this

color(blue)("pOH" = pK_b + log( (["conjugate acid"])/(["weak base"])))

The base dissociation constant, ${K}_{b}$, for ethylamine is listed as being equal to $5.6 \cdot {10}^{- 4}$

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

Now, you know that this buffer solution is 1.40% ethylamine and 1.18% ethylammonium bromide by mass.

Let's assume that you have a sample of volume $V$ and mass $m$ of this buffer solution. In order to be able to find the concentrations of the weak acid and of its conjugate base, you need to first determine how many moles of each you have.

Use the given percent concentrations to determine how many grams of each you'd have

m color(red)(cancel(color(black)("g solution"))) * ("1.40 g C"_2"H"_5"NH"_2)/(100color(red)(cancel(color(black)("g solution")))) = 1.40/100 * m color(white)(x)"g C"_2"H"_5"NH"_2

m color(red)(cancel(color(black)("g solution"))) * ("1.18 g C"_2"H"_5"NH"_3"Br")/(100color(red)(cancel(color(black)("g solution")))) = 1.18/100 * m color(white)(x)"g C"_2"H"_5"NH"_3"Br"

Now use the molar masses of the weak base and the salt of its conjugate acid to determine how many moles of each you have

1.40/100*mcolor(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_5"NH"_2)/(45.08color(red)(cancel(color(black)("g")))) = (0.0311m)/100"moles C"_2"H"_5"NH"_2

and

1.18/100*mcolor(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_5"NH"_3"Br")/(126.0color(red)(cancel(color(black)("g")))) = (0.00937m)/100"moles C"_2"H"_5"NH"_3"Br"

Now, since ethylammonium bromide dissociates to produce the ethyammonium ion in a $1 : 1$ mole ratio, it follows that the solution will contain $\frac{0.00937 m}{100}$ moles of ethylammonium ions.

Now plug these values into the H-H equation and solve for the $\text{pOH}$ of the solution

$\text{pOH} = - \log \left(5.6 \cdot {10}^{- 4}\right) + \log \left(\frac{\frac{0.00937 \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}}}{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{100}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{\frac{0.0311 \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}}}{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{100}}}}}\right)$

$\text{pOH} = 3.25 + \log \left(\frac{0.00937}{0.0311}\right) = 2.73$

This means that the pH of the solution will be

$\text{pH" = 14 - "pOH}$

$\text{pH} = 14 - 2.73 = \textcolor{g r e e n}{11.27}$