# 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?

##### 1 Answer

#### Answer:

#### Explanation:

You're dealing with a buffer solution that contains *ethylamine*, *ethylammonium bromide*, *ethylammonium ion*,

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*,

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

Now, you know that this buffer solution is **by mass**.

Let's assume that you have a sample of volume *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

Now plug these values into the H-H equation and solve for the

#"pOH" = -log(5.6 * 10^(-4)) + log( ( (0.00937color(red)(cancel(color(black)(m))))/color(purple)(cancel(color(black)(100))))/color(red)(cancel(color(black)(V))) * color(red)(cancel(color(black)(V)))/((0.0311color(red)(cancel(color(black)(m))))/color(purple)(cancel(color(black)(100)))))#

#"pOH" = 3.25 + log(0.00937/0.0311) = 2.73#

This means that the pH of the solution will be

#"pH" = 14 - "pOH"#

#"pH" = 14 - 2.73 = color(green)(11.27)#