Question

Zorcaine, C13H24N,SCI, is the salt of the base articaine and hydrochloric acid. The ionization constant for articaine is 6.4 x 10-7. (a) Is a solution of zorcaine acidic or basic? acidic O basic (b) What are the [H20+], [OH-] (in M), and pH of a 2.0% solution by mass of zorcaine, assuming that the density of the solution is 1.0 g/mL. (Assume Kw = 1.01 10-14.) [H30+1 4.9) х м [OH^] 49 M 4.0 pH

Answer 1

(a) Is a solution of zoracaine acidic or basic? : acidic

(b) [H₃O⁺] = 3.14 x 10^-5 M

[OH^-] = 3.22 x 10^-10 M

pH = 4.50

Explanation

Let mass of solution = 1000 g

mass of zoracaine = (mass fraction zoracaine) * (mass of solution)

mass of zoracaine = (2.0 / 100) * (1000 g)

mass of zoracaine = 20 g

moles zoracaine = (mass zoracaine) / (molar mass zoracaine)

moles zoracaine = (20 g) / (320.8354 g/mol)

moles zoracaine = 0.0623 mol

concentration zoracaine = (moles zoracaine) / (volume of solution in Liter)

concentration zoracaine = (0.0623 mol) / (1.0 L)

concentration zoracaine = 0.0623 M

K_b articaine = 6.4 x 10^-7

K_a zoracaine = (K_w) / (K_b)

K_a zoracaine = (1.01 x 10^-14) / (6.4 x 10^-7)

K_a zoracaine = 1.578 x 10^-8

ICE table	C13H21N2O3S+ (aq)	$\rightleftharpoons$	C13H20N2O3S (aq)	H+ (aq)
Initial conc.	0.0623 M		0	0
Change	-x		+x	+x
Equilibrium conc.	0.0623 M - x		+x	+x

K_a = [C₁₃H₂₀N₂O₃S]_eq[H⁺]_eq / [C₁₃H₂₁N₂O₃S⁺]_eq

1.578 x 10^-8 = [(x) * (x)] / (0.0623 M - x)

Solving for x, x = 3.14 x 10^-5 M

[H⁺] = x = 3.14 x 10^-5 M

pH = -log[H⁺]

pH = -log(3.14 x 10^-5 M)

pH = 4.50