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

Question

Serenity Allen · Accepted Answer

#"pH" = 11.27#

You're dealing with a buffer solution that contains ethylamine, #"C"_2"H"_5"NH"_2#, a weak base, and ethylammonium bromide, #"C"_2"H"_5"NH"_3"Br"#, the salt of its conjugate acid, the ethylammonium ion, #"C"_2"H"_5"NH"_3^(+)#.

This is the Henderson-Hasselbalch equation for a conjugate acid buffer with a weak base.

#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 * 10^(-4)#

Table of Weak Bases: https://tutor.hix.ai

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.

Utilize the specified percentage concentrations to calculate the grams that you would have of each.

#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, calculate how many moles of each you have using the molar masses of the weak base and the salt of its conjugate acid.

#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#

additionally

#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 #(0.00937m)/100# moles of ethylammonium ions.

Now plug these values into the H-H equation and solve for the #"pOH"# of the solution

#"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 implies that the solution's pH will be

#"pH" = 14 - "pOH"#

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

Avery Adams · Answer

undefined You would use the Henderson-Hasselbalch equation: pH = pKa + log([A^-]/[HA]). 
1. Calculate the concentrations of the weak base (C2H5NH2) and its conjugate acid (C2H5NH3Br) in the solution.
2. Convert the mass percentages to grams.
3. Calculate the moles of each compound using their molar masses.
4. Determine the volume of the solution.
5. Use the concentrations and volume to calculate the molarity of each compound.
6. Use the Henderson-Hasselbalch equation with the pKa value for the C2H5NH2/C2H5NH3^+ pair to find the pH of the solution.