In: Chemistry
Suppose that you have 0.500 L of each of the following solutions, and an unlimited supply of water. Kb= 6.3×10−5for trimethylamine, (CH3)3Nand Ka= 1.9×10−5for hydrazoic acid, HN3. 0.114mol L−1(CH3)3N(aq) 0.124 mol L−1HN3(aq) 0.102mol L−1HI(aq) 0.113mol L−1KOH(aq) Provide simple instructions for preparing 1.00 L of a buffer solution having pH = 9.00 at 298 K. Focus on preparing the buffer that containsthe highest possible concentrations of the active components. Your instructions should include the volumes of the solutions require
V = 1 L of buffer with pH = 9
first, identify pK's values:
pKb triethyl amine = 6.3*10^-5 --- >-log( 6.3*10^-5) = 4.2
pKa Hydrazoic acid = 1.9*10^-5 --> -log(1.9*10^-5) = 4.721
[B] = 0.114 M
[A] = 0.124 M
HI and KOH are also present
now... the best will be to get the buffer near pH = pKa vlaue
pOH = 14-pH = 14-9 = 5
this best suites the base with pKb = 4.2
so..
pOH = pKb + log(HB+ / B)
5= 4.20 + log(HB+ / B)
[HB+]/[B] = 10^(5-4.20) = 6.309
[HB+] = 6.30[B]
for 1 liter solution:
HB+ moles = 6.3*B moles
so..
initially
mol of B = M1V1 = x
mol of HB+ = 0
after reaction
mol of B = M1V1 - Macid*Vacid = 0.114*V1 - 0.102*Vacid
mol of HB+ = 0+Macid*Vacid = 0.102*Vacid
also
HB+ moles = 6.3*B moles
0.102*Vacid = 6.3(0.114*V1 - 0.102*Vacid)
0.102*Vacid = 0.7182*V1 - 0.643*Vacid
(0.102+0.643) * VAcid = 0.7182*V1
so..
V1+Vacid = 1L
Vacid = 1-V1
(0.102+0.643) * (1-V1)= 0.7182*V1
0.745 - 0.745V1 = 0.7182*V1
(0.7182+0.745)*V1 = 0.745
V1 = (0.745) / (0.7182+0.745) = 0.5091 liters
Vacid = 1-0.5091 = 0.4909liters