200 likes | 222 Views
Learn about acids, bases, pH, and pOH measurements with examples and calculations. Discover the relationship between hydrogen and hydroxide ion concentrations.
E N D
Acids and Bases pH and pOH
Intro Quiz: • Put these items in pH order from most acidic to least. • Blood • Milk • Water • Lemon juice • Baking soda • Coffee • Ammonia
Answer: • Lemon juice • Coffee • Milk • Water • Blood • Ammonia • Baking Soda
pH • pH is a measure of the hydrogen ion concentration, [H+] • pH is calculated using the following formula:pH = -log10[H+]
pOH • pOH is a measure of the hydroxide ion concentration, [OH-] • pOH is calculated using the following formula:pOH = -log10[OH-]
Solutions • Acidic solution: When the [H+] is greater than [OH-] • Basic solution: When [OH-] is greater than [H+] • Basic solutions are also known as alkaline.
Example 1: • Find the pH of a 0.2mol L-1 (0.2M) solution of HCl • Calculate pH: pH = -log10[H+]pH = -log10[0.2] = 0.7
Example 2: • Find the pOH of a 0.1mol L- (0.1M) solution of NaOH • Calculate pOH: pOH = -log10[OH-]pOH = -log10[0.1] = 1
Example 3: • Find the pH of a 0.2 mol L-1 (0.2M) solution of H2SO4 • Calculate pH: pH = -log10[H+]pH = -log10[0.4] = 0.4
Example 4: • Find the pOH of a 0.1mol L-1 (0.1M) solution of Ba(OH)2 • Calculate pOH: pOH = -log10[OH-]pH = -log10[0.2] = 0.7
pH + pOH = 14 • And: • [H+] * [OH-] = 1 x 10-14
OR: • [H+]*[OH-] = Kw = 1 X 10-14 • Take the -log of both sides of the equation • - log [H+] +(- log [OH- ]) = - log [1 X 10-14 ] • pH + pOH = 14
About those H+’s • Water molecules are highly polar • A water molecule that loses H+ becomes the negatively charge hydroxide ion OH- • The water molecule that gains a H+ (proton) becomes a hydronium ion, H30+ • H2O (l) H3O+(aq) + OH-(aq)
Self Ionization • This is also known as the self ionization reaction of water • H2O (l) H3O+(aq) + OH-(aq) • This is because water is amphoteric.
Example (1): • Find the pH of a solution of sodium hydroxide that has a pOH of 2 • pH = 14 - pOH • pH = 14 - 2 = 12
Example (2): • Find the [OH-] of a sulfuric acid solution with a pH of 3 • Calculate the pOH • pOH = 14 - pH • pOH = 14 - 3 = 11 • Calculate [OH-][OH-] = 10-pOH[OH-] = 10-11 = 10-11 mol L-1
Example • pH [H+] [OH-] pOH • 7 1.0 x 10-7M 1.0 x 10-7M 7 • 6 1.0 x 10-6M 1.0 x 10-8M 8 • 5 1.0 x 10-5M 1.0 x 10-9M 9 • Etc.