Question

Complete the following table for a selection of foods:

Food	[H3O+]	[OH}	pH	Acidic, basic, or neural
Apple Juice			3.80
Soy Milk				neutral
Tuna (canned)	6.8 x 10^-7

Part A: calculate the [H3O+] of apple juice with pH = 3.80 (express the molarity to two significant figures)

Part B: calculate the [OH] of apple juice with pH = 3.80 (express the molarity to two significant figures)

Part C: calculate the [H30+] of soy milk (express the molarity to two significant figures)

Part D: calculate the [OH] of soy milk (express the molarity to two significant figures)

Part E: calulate the pH of soy milk (express the pH to two decimal places)

Part F: calculate the [OH] of tuna (canned) with [H3O+]= 6.8 x10^-7 (express the molarity to two significant figures)

Part G: calculate the pH of tuna (canned) with[H30+] = 6.8x 10^-7 (express the pH to two decimal places)

Answer 1

part A

pH = 3.8

[H3O+] = 10^-3.8 = 1.585*10^-4 M

part B

pH = 3.8

[H3O+] = 10^-3.8 = 1.585*10^-4 M

KW = 1*10^-14 = [H3O+][OH-]

[OH-] = 1*10^-14 / ( 1.585*10^-4) = 6.3*10^-11 M

part C

for neutral solution [H3O+] = [OH-]

KW = 1*10^-14 = [H3O+][OH-]

SO that

[H3O+] = 1*10^-7 M

part d

for neutral solution [H3O+] = [OH-]

KW = 1*10^-14 = [H3O+][OH-]

SO that

[OH-] = 1*10^-7 M

part E

pH = -log(H3O+)

    = -log(1*10^-7)

   = 7.0

part F

[OH-] = KW/[H3o+] = 1*10^-14 / (6.8*10^-7)

       = 1.47*10^-8 M

part G

pH = -log(6.8*10^-7)

   = 6.2