Walter White, aka Heinsenberg, created a titration curve for methamphetamine (wh

Question

Walter White, aka Heinsenberg, created a titration curve for methamphetamine (which is a weak base, pKa 9.87) using 0.01M HCl. He titrated 50.0 mL of a 6.7 x10-3 M solution of methamphetamine with HCl. Prepare a titration curve based on the following points, sketch your curve on below graph, label the half equivalence point and equivalence points on the graph. Show all work

Initial pH, 0.0 mL of HCl

After 10.0 mL of HCl

After 16.8 mL of HCl

After 25.0 mL of HCl

After 30.0 mL of HCl

After 33.5 mL of HCl

After 40.0 mL of HCl

After 50.0 mL of HCl

Explanation / Answer

The no. of mmol of methamphetamine = 50 mL * 6.7 *10-3 mmol/mL = 0.335 mmol

(i) Volume of acid = 0 mL

pOH = 1/2 (pKb - Log[methamphetamine])

= 1/2 (14 - 9.87 - Log(6.7 *10-3))

= 3.15

Therefore, pH = 14-3.15 = 10.85

(ii) The volume of HCl = 10 mL

The no. of mmol of HCl = 10 mL * 0.01 mmol/mL = 0.1 mmol

Here, 0.1 mmol of HCl reacts with 0.1 mmol of methaphetamine to form 0.1 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.1 mmol

The no. of mmol of unreacted methamphetamine (nmethamphetamine) = 0.335-0.1 = 0.235 mmol

According to Henderson-Hasselbulch equation:

pH = pKa + Log([nmethamphetamine]/nmethamphetamine.HCl])

= 9.87 + Log(0.235/0.1)

= 10.24

(iii) The volume of HCl = 16.8 mL

The no. of mmol of HCl = 16.8 mL * 0.01 mmol/mL = 0.168 mmol

Here, 0.168 mmol of HCl reacts with 0.168 mmol of methaphetamine to form 0.168 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.168 mmol

The no. of mmol of unreacted methamphetamine (nmethamphetamine) = 0.335-0.168 ~ 0.168 mmol

According to Henderson-Hasselbulch equation:

pH = pKa + Log([nmethamphetamine]/nmethamphetamine.HCl])

= 9.87 + Log(0.168/0.168)

= 9.87

Since pH = pKa, this point (Volume of HCl = 16.8 mL) is the half-equivalence point.

(iv) The volume of HCl = 25 mL

The no. of mmol of HCl = 25 mL * 0.01 mmol/mL = 0.25 mmol

Here, 0.25mmol of HCl reacts with 0.25 mmol of methaphetamine to form 0.25 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.25 mmol

The no. of mmol of unreacted methamphetamine (nmethamphetamine) = 0.335-0.25 = 0.085 mmol

According to Henderson-Hasselbulch equation:

pH = pKa + Log([nmethamphetamine]/nmethamphetamine.HCl])

= 9.87 + Log(0.085/0.25)

= 9.40

(v) The volume of HCl = 30 mL

The no. of mmol of HCl = 30 mL * 0.01 mmol/mL = 0.3 mmol

Here, 0.3 mmol of HCl reacts with 0.3 mmol of methaphetamine to form 0.3 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.3 mmol

The no. of mmol of unreacted methamphetamine (nmethamphetamine) = 0.335-0.3 = 0.035 mmol

According to Henderson-Hasselbulch equation:

pH = pKa + Log([nmethamphetamine]/nmethamphetamine.HCl])

= 9.87 + Log(0.035/0.3)

= 8.94

(vi) The volume of HCl = 33.5 mL

The no. of mmol of HCl = 33.5 mL * 0.01 mmol/mL = 0.335 mmol

Here, 0.335 mmol of HCl reacts with 0.335 mmol of methaphetamine to form 0.335 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.335 mmol

Here, nmethamphetamine.HCl = nmethamphetamine

Therefore, this point (Volume of HCl = 33.5 mL) is the equivalence point.

i.e. [nmethamphetamine.HCl] = 0.335 mmol/(33.5 + 50) mL = 0.004 M

Now, pH = 7 - 1/2 (pKb + Log[nmethamphetamine.HCl])

= 7 - 1/2 {4.13 + Log(0.004)}

= 6.13

(vii) The volume of HCl = 40 mL

The no. of mmol of HCl = 40 mL * 0.01 mmol/mL = 0.4 mmol

Here, 0.335 mmol of HCl reacts with 0.335 mmol of methaphetamine to form 0.335 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.335 mmol

The no. of mmol of unreacted HCl = 0.4-0.335 = 0.065 mmol

[HCl] = [H+] = 0.065 mmol/(50+40) mL = 7.22*10-4 M

pH = -Log[H+] = -Log(7.22*10-4 M) = 3.14

(viii) The volume of HCl = 50 mL

The no. of mmol of HCl = 50 mL * 0.01 mmol/mL = 0.5 mmol

Here, 0.335 mmol of HCl reacts with 0.335 mmol of methaphetamine to form 0.335 mmol of methamphetamine.HCl. i.e. nmethamphetamine.HCl = 0.335 mmol

The no. of mmol of unreacted HCl = 0.5-0.335 = 0.165 mmol

[HCl] = [H+] = 0.165 mmol/(50+50) mL = 1.65*10-3 M

pH = -Log[H+] = -Log(1.65*10-3 M) = 2.78

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