Suppose that you have 0.500 L of each of the following solutions, and an unlimit

Question

Suppose that you have 0.500 L of each of the following solutions, and an unlimited supply of water. Kb= 6.3×105for trimethylamine, (CH3)3Nand Ka= 1.9×105for hydrazoic acid, HN3. 0.114mol L1(CH3)3N(aq) 0.124 mol L1HN3(aq) 0.102mol L1HI(aq) 0.113mol L1KOH(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

Explanation / Answer

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

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