CHM 1025C Module Eight-Chapter 15.8-15.9 Name:__Answers___

Part L: pH Scale Calculations (Section 15.8-15.9) 10 points

What is the pH and pOH of the following solutions:

pOH = 14.00 – pH K_w = 1.00 x 10^-14 = [H₃O¹⁺][OH^1-] [H₃O¹⁺] is the same as [H¹⁺]

(1) 0.0100 M HCl

pH = -Log [H⁺] = -Log [0.0100] = 2.00

pOH = 14.00 – pH = 14.00 – 2.00 = 12.00

(2) 0.0055 M Mg(OH)₂

The dissociation reaction: Mg(OH)₂ ↔ Mg ²⁺ + 2 OH ^1-

A 0.0055 M Mg(OH)₂ solution produces a 2 x 0.0055 M hydroxide ion or 0.0110 M hydroxide ion

pOH = -Log [OH^-] = -Log [0.0110] = 1.96

pOH = 14.00 – pH = 14.00 – 1.96 = 12.04

(3) The pH of vinegar is 2.85, calculate the [H₃O¹⁺]:

[H⁺] = 10^–pH = 10^-2.85 = 0.00141 M

(4) Easy Off Oven Cleaner has a pH of 11.70. What is the [H₃O¹⁺] and [OH ^1- ]?

[H⁺] = 10^{–pH =} 10^-11.70 = 1.995 x 10^-12 M

pOH = 14.00 – pH = 14.00 – 11.70 = 2.30

[OH^-] = 10^{–pOH =} 10^-2.30 = 0.00501 M