Question

please answer all parts A- J

my school shut down due to virus and i cant get tutoring anymore. ive veeb stuck on this for over an hour

<Assignment 9 (Chapter 15) Problem 15.19 Acetic acid, CH3CO2H, is the solute that gives vinegar its characteristic odor and sour taste.

Answer 1

Part A

Acetic acid is weak acid

It dissociates as

CH3COOH (aq) $\to$ CH3COO- (aq) + H+ (aq)

Let concentration of acetic acid = c

Degree of dissociation is x

ICE table is

Concentration	CH3COOH	CH3COO-	H+
Initial	c	0	0
Change	-cx	+cx	+cx
Equilibrium	c-cx	cx	cx

Acid dissociation constant (Ka)

Ka =

CH3C00-[H+] CH3COOH

Or, Ka = (cx × cx/(c - cx)

Or, Ka = cx2/(1-x)

As, x <<<1, so, Ka = cx2

Or, x = √(Ka/c)

or, cx = √(Ka.c)

Or, [H3O+] = Cx = √(Ka.c)

pKa = 4.75

Hence, formula for pH calculation is

pH = $\frac{1}{2}$ ( pKa - logCacid)

Given , Concentration of acetic acid (Cacid)

= 1.24 M

Or, pH = $\frac{1}{2}$ ( 4.75 - log1.24)

= 2.42.

Part B

pH = - log[H3O+]

Or, 2.42 = - log [H3O+]

Or, [H3O+] = 10(-2.42) = 0.0038 M.

Part C

cx = [CH3CO2 -] = [H3O+] = 0.0038 M.

Part D

[CH3CO2H] = C - Cx

= (1.24 - 0.0038) M

= 1.2362 M.

= 1.2 M (2 significant figures)

Part E.

[OH-] = 10-14/[H3O+]

Or, [OH-] = [10-14/ (0.0038)]

= 2.6×10-12 M.

Part F.

pH = $\frac{1}{2}$ (pKa - logC)

= $\frac{1}{2}$ ( 4.75 - log0.0158)

= 3.28

Part G

[H3O+] = 10(-3.28) = 0.00052 M

Part H

[CH3CO2-] = 0.00052 M.

Part I

[CH3CO2H] = (0.0158 - 0.00052) M

= 0.015 M ( 2 significant figure).

Part J.

[OH-] = [10-14/0.00052]

= 1.9× 10-11 M