Question

All parts of the hemlock plant contain the alkaloid coniine. Coniine (C8H17N) is an alkaline liquid with a pKb of 1 and a density of 0.844 g/ml. Ingestion of the substance may cause weakness, drowsiness, nausea, vomiting, labored respiration, paralysis and death. If an attempted murderer poured 2.5 ml of coniine into someone’s 500 ml water bottle, what would the pH be for the resulting solution? If the lethal dose for mice of 7 mg/kg of body weight is similar to that for humans, would a 120 pound athlete die from drinking the contents of the water bottle?

Answer 1

pH of resulting solution = 12.42

would a 120 pound athlete die? : Yes

Explanation

volume coniine = 2.5 mL

mass coniine = (volume coniine) * (density coniine)

mass coniine = (2.5 mL) * (0.844 g/mL)

mass coniine = 2.11 g

moles coniine = (mass coniine) / (molar mass coniine)

moles coniine = (2.11 g) / (127.2273 g/mol)

moles coniine = 0.01658 mol

concentration coniine = (moles coniine) / (total volume in Liter)

concentration coniine = (0.01658 mol) / (0.5025 L)

concentration coniine = 0.0330 M

pK_b = 1

K_b = 10^-pK_b

K_b = 10^-1

K_b = 0.10

ICE table	C8H17N (aq)	H2O (l)		C8H17NH+ (aq)	OH- (aq)
Initial conc.	0.0330 M	-		0	0
Change	-x	-		+x	+x
Equilibrium	0.0330 M - x	-		+x	+x

K_b = [C₈H₁₇NH⁺]_eq[OH^-]_eq / [C₈H₁₇N]_eq

0.10 = [(x) * (x)] / (0.0330 M - x)

Solving for x, x = 0.02616 M

[OH^-] = x = 0.02616 M

pOH = -log[OH^-]

pOH = -log(0.02616 M)

pOH = 1.58

pH = 14 - pOH

pH = 14 - 1.58

pH = 12.42

(b.) mass of athlete = 120 lb

mass of athlete = 120 lb * (1 kg / 2.20462 lb)

mass of athlete = 54.43 kg

lethal amount of coniine = (mass of athlete) * (lethal dose)

lethal amount of coniine = (54.43 kg) * (7 mg/kg)

lethal amount of coniine = 381.0 mg

lethal amount of coniine = 0.381 g

Since mass of coniine injested (2.11 g) is greater than lethal amount (0.381 g), therefore, athlete will die.