Acid Base Equilibrium

• Why can we eat and drink some acids and bases and not others?
• The degree to which they release or accept protons
• Equilibrium constants describe how readily acids deprotonate and bases protonate.

Self-Ionization of Water

Water can react with itself.

One molecule accepts and one donates a proton. 

This is called self-ionization

• This reaction is very important. Because it gives us a constant for acid base equilibrium. Kw is the equilibrium constant for water. 
• Kw = [OH^-][H₃O⁺] = 1.0 x 10^-14 M
• Note: M stands for molarity which is moles/L
• Whether a solution is acidic, basic or neutral, the product of the [H3O+] and [OH-] is always equal to Kw.
• Example problem: 
• The [H₃O⁺] in a mild acid is found to be 5 x 10^-7 mol/L. What is the concentration (molarity) of hydroxide ions?
• [OH^-][H₃O⁺] = 1.0 x 10^-14 M
• [OH^-] = 1.0 x 10^-14 M/[H₃O⁺]
• [OH^-] = 1.0 x 10^-14 M/5.0 x 10^-7 M
• [OH^-] = 2.0 x 10^-8 M

pH Scale

• pH stands for “power of hydronium”
• pH is the negative logarithm of the [H₃O⁺]
• pH = -log [H₃O⁺]
• Examples:
• If [H₃O⁺] = 0.0025 M
• pH = -log(0.0025) = 2.6
• If [H₃O⁺] = 4.57 x 10^-9 M
• pH = -log(4.57 x 10^-9 M) = 8.34

pH Values

• A pH of 7 means neutral.
• A pH of 0-7 is an acid.
• A pH of 7-14 is basic

pH Scale

• Example 1:
• The [H₃O⁺] in a shampoo is 2.0 x 10^-5 M. What is the pH of this shampoo?
• pH = -log [H₃O⁺]
• pH = -log(2.0 x 10^-5) = 4.7
• Example 2:
• What is the pH of an aqueous solution of 0.40 g of HI dissolved in 500 mL of water?
• First, convert grams of HI to moles of HI.
• 0.4 g HI x (1 mol HI/127.9 g HI) = 0.00031 mol HI
• Next calculate molarity (M).
• (0.0031 mol HI/500 mL) x (1000 mL/1L) = 6.3 x 10^-3 M 
• Solution:
• pH = -log[H₃O⁺
•       = -log(6.3 x 10^-3 M)
•       =  2.2
• Example 3
• Find the pH of a solution whose [H₃O⁺] equals 9.5 x 10^-8.
• pH = -log[H₃O⁺]
•       = -log(9.5 x 10^-8 M)
•       =  7.02