Question

We started with 200.0mL of an 0.08 M Solution of the sodium salt of valine. We then added 25.0mL of 1.00 M HBr. pka1= 2.286 pka2= 9.719 Ka1= 5.18x10^-3 Ka2= 1.91x10^-10 Please explain the solution and where it is based on the equivalence points!

a) What is the new solution made? Be descriptive.

b) What is the pH of the new solution?

Answer 1

a)

First, let us anlayse the composition of the new solution

Total Volume = 200 + 25 = 225 mL

volume has increased

now,

mmol of sodium salt of valine = M*V = 0.08*200 = 16 mmol of sodium salt

then, calcualte acid added

mmol of HBr= MV = 25*1 = 25 mmol of acid

clearly, the sodium salt will reac twith the extra acid

Sodium Salt of Valine can be represented as Na2*Va

where Va- is the conjugate base of the valine acid

so

Va-2+ HBr = Br- + HVa-

there will be HBa and HBr

mmol of HBr added = 25

mmol of Va-2 reacted = 16

after reaction:

mmol of HBr reacted = 25-16 = 9

mmol of HVa- formed = 0 + 16 = 16

note that this was for the first proton, now we hav ethe second proton

after more HBr is present:

HBr + HVa- = H2Va + Br-

mmol of HVa- initially = 16

mmol of HBr initially = 9

after reaction

mmol of H2Va present = 9

mmol of HVa- left = 16-9 = 7

this is now a buffer

since we have H2Va and HVa-

this is the 2nd ionization so we need pK2

for a buffer, apply henderson hasselbach equation

pH = pKa2 + log(HVa-/H2VA)

pH = 9.719+ log(7/9)

pH = 9.609855