Warm Up 4/14
How many protons and neutrons would an H+
ion have?
Acids Bases
Definitions
Neutralization
Strong vs Weak
Litmus Paper
Intro to Acids & Bases
Common Acids
Properties of Acids- Corrosive (burns skin and “eats” metals)
- Sour taste- Turns blue litmus paper red- Electrolyte-pH less than 7
Common Bases
Properties of Bases-Caustic (leaves a white residue on metals)-Bitter taste-Slippery feeling-Turns red litmus blue-Electrolyte-pH greater than 7
ArrheniusAcid: a solution made of any solute with hydrogen (H+) as its cation.Base: a solution of any ionic salt with hydroxide (OH-) as its anion.
BronstedAcid: molecule or ion that is a
proton (H+) donorBase: a molecule or ion that is a
proton acceptor
Neutralization Reaction – When an acid and base react to form a neutral solution (water and a salt).Examples:• NaOH + HCl → H2O + NaCl• 2NaOH + H2SO4 → 2H2O + Na2SO4
Acids and bases can be classified as weak or strong.
Strong acid/base- ionizes (dissociates) completely
ex.: HCl + H2O → H+(aq) + Cl-(aq)
Weak acid/base- only ionizes (dissociates) partially
ex.: HF + H2O → H+(aq) + F-(aq) + HF
Warm Up 3/26
Give the salt that will be formed in the following neutralization reaction:
2 HBr + Ba(OH)4 → 2 H2O + ___?___
•pH stands for “power of hydronium”•Hydronium is H3O+ (it is the correct way to write the formula of a H+ ion that has been donated in water)
pH Scale
•Acids have a pH less than 7• 7 is neutral• Bases have a pH higher than 7•Each step on the pH scale is a step of 10
Calculating pH
• pH or “power of hydronium” is based on the concentration of H3O+ ions
• H3O+ is another way to write H+
• Formula: pH = -log [H+]
Calculator Steps
1)Press the (-) button (not the minus sign button!)
2)Press the Log button3)Enter the concentration and
close the parenthesis
Example #1Calculate the pH of a solution
whose [H+] = 1.0 x 10-4
Example #2 Calculate the pH of a solution
whose [H+] = 3.1 x 10-4
Left Side PracticeFind the pH of these concentrations of H+, and tell whether it’s an acid or a base.1) [H+] = 1.0 ×10−6 M 2) [H+] = 2.19 ×10−4 M 3) [H+] = 9.18 ×10−11 M 4) [H+] = 4.71 ×10−7 M 5) [H+] = 1.0 M
K = [H+][OH-] = 1.0 ×10−14 w
Ionization Constant of Water:
(We will be using this in our calculations)
Finding pH from [OH-] •Divide 1.0 ×10−14 by the [OH-] concentration.
•Then solve like a regular H+ problem: -log [H+]
Example: Calculate the pH of a solution with an [OH-] of 1.0 ×10−3 .
pOH CalculationsFormula: pOH = -log [OH-]
• If given the [OH-], then just put that number in the parenthesis.
• If given the [H+], then divide it by 1.0 ×10−14 and put the answer in the parenthesis.
Example: Calculate the pOH of a solution with an [OH-] of 2.7×10−4 .
Example: Calculate the pOH of a solution with an [H+] of 9.18 ×10−11 .
To go from pH back to [H+] (or from pOH back to [OH-]) just put in: 10^- given
Example: Calculate the [H+] of a solution whose pH = 6.3
Calculating pOH
pOH or “power of hydroxide ion”Formula:
pOH = -log [OH-]
Example #3
What is the pH of a 0.250M solution of KOH?
Example Problem #4
Calculate the pH and pOH of a 0.315M H2SO4 solution.