Question

You are working for the Crime Lab and you habe been tasked with determining how much of the weak base cocaine (k_b =2.6 x 10^-6) is present in a sample of white powder. you dissolve 5.0 g of the white powder into 50.00 mL of water. You add.26.00 mI, of 0.025 M HCI titrant to reach the equivalence point.

a.) what is the pH of solution at the equivalence point?  what was the pH of your original 50.00 mL solution?

Answer 1

You are working for the Crime Lab and you habe been tasked with determining how much of the weak base cocaine (k_b =2.6 x 10^-6) is present in a sample of white powder. You dissolve 5.0 g of the white powder into 50.00 mL of water. You add.26.00 mI, of 0.025 M HCI titrant to reach the equivalence point.

Number of base = 0.00065 moles

Now molarity of base = 0.00065/ 0.050

= 0.013M

a.) what is the pH of solution at the equivalence point?  what was the pH of your original 50.00 mL solution?

At the equivalence point, you have moles of HCl equal to the moles of weak base that we began with. So, in the resulting solution, you have just the protonated form of weak base, BH+. The concentration of this can be calculated as follows:

Total volume = 50.00+26.00 = 76.00 ml = 0.076 L

Number of moles = moalrity * volume in L

= 0.025*0.026

= 0.00065 moles of HCl

Molarity of weak base =0.00065 / 0.076 = 0.0085 M

Now, the BH+ ion ionizes as an acid by the equation:

BH+ + H2O <---> B + H3O+

Ka = [B][H3O+]/[BH+]

Given that; k_b =2.6 x 10^-6

The value of Ka is calculated from Kb as:
KaKb = Kw
Ka = 1.00X10^-14 / Kb ; 2.6X10^-6

= 3.8 5 x 10^-9

Therefore;

Ka = [B][H3O+]/[BH+]

3.8 5 x 10^-9= x*x / 0.0085 –x

Assume that x will be small compared to 0.0085

Then;

3.8 5 x 10^-9= x^2 / 0.0085

x = 5.72X10^-6 = [H3O+]/

pH = -log [H+]

= - log (5.72X10^-6)

= 5.24

what was the pH of your original 50.00 mL solution?

[BASE ]= 0.013M

B + H2O = BH + + OH-

K b [BH+][OH-]/ [B]

k_b =2.6 x 10^-6

k_b = X*X / 0.013-x

Assume that x will be small compared to 0.013

Then;

2.6 x 10^-6 = X*X / 0.013

X= 1.84*10^-4 = [OH-]

pOH = - log [OH-]

= - log 1.84*10^-4

= 3.74

pH + pOH = 14

pH = 14-3.74

= 10.26