Chemistry Guide: The Quick Guide to Acids and Bases
by David Phelps, Founder
david@forumeducation.nyc
BASIC PRINCIPLES
To neutralize (titrate) a solution, the following must be true:
- moles of H+ = moles of OH-
To get there? This is the role of acids and bases.
ARRHENIUS DEFINITION (older, for aqueous solutions):
- Acid: Produces Hydrogen ions (H+)
- Base: Produces Hydroxide ions (OH-)
BRONSTED-LOWRY DEFINITION (modern, more general):
- Acid: Gives hydrogen ions (H+)
- Base: Takes hydrogen ions
- Conjugate acid: Given hydrogen ions
- Conjugate base: From which hydrogen ions have been taken
Note: The acid and conjugate acid are not based on the same compound
RECOGNIZING ACIDS AND BASES (WORKFLOW)
Note: Amphiprotic molecules (like H2O and HCO3-) can take or give H+ ions, so use flow chart to determine whether they act in a particular reaction as acids or bases.
MEASURING ACID AND BASE STRENGTH
For Binary Acids
In a binary acid, a single atom bonds to the hydrogen ion. In other words, given the form H+A-, A- is a single atom.
Electronegativity Rules
- Horizontally, rightward anions are electronegative
- Electronegative ions draw electrons to one side of the acid, polarizing the molecule and leaving H+ on one side.
- Rule #1: The farther right the anion, the stronger the acid.
Bond Strength Rules
- Vertically, less bond strength means stronger acids.
- When bonds are weak, it’s easier to break off an H+.
- Rule #2: The lower the anion, the stronger the acid.
Polyatomic Acids
In polyatomic acids, polyatomic ions bond to the hydrogen ion. Given the form H+AO#-, AO#- is a polyatomic ion (almost always with oxygen).
Electronegativity Rules
- Vertically, higher anions draw electrons from H+
- Rule #3: the higher a polyatomic acid’s central atom, the stronger it is
- Bond strength doesn’t matter: central atom doesn’t bond w/ H+
Additional Oxygens
- Each new O draws electrons from the “OH” bond
- Rule #4: The more Os, the stronger the acid.
Trick: Use electronegativity unless you’re comparing binary acids vertically
Bases
To determine base strength, just reverse all the rules above.
DISSOLVING SALTS IN WATER
Two Definitions
- A salt is formed when an acid and base neutralize each other.
- Salts are composed of the cation of a base and an anion of an acid.
Acidic salts
Acidic salts are formed from strong acids and weak bases. They give up their H+ ion to water and form a base as a result. In other words:
- Acidic salt solutions produce weak bases
- Acidic salt solutions are acidic (pH < 7)
Basic salts
Basic salts are formed from strong bases and weak acids. They take a H+ from water and form an acid as a result, while leaving a OH- behind. So:
- Basic salt solutions produce weak acids
- Basic salt solutions are basic (pH > 7)
For Example
What happens when we dissolve NaCH3COO in water?
- NaCH3COO was formed from a strong base (NaOH) and weak acid (HCH3COO), so it’s a basic salt.
- It won’t produce NaOH in H2O since NaOH is a strong base: the ions dissolve.
- But it can react with water to produce HCH3COO, which is weak.
The key points?
- A basic salt produces a weak acid (HCH3COO)
- A basic salt results in a basic solution since it produces OH-
GENERAL STRATEGIES FOR DETERMINING pH
pH = -log[H+]
pOH = -log[OH-]=14-pH
- High concentrations of H+= a low pH = very acidic.
- High concentrations of OH- = a low pOH = a high pH = very basic
Strong Acid or Base Solution
For example, HCl
- Calculate total moles of H+ or OH- ions
- Plug into pH = -log[H+] or pOH = -log[OH-]=14-pH
Strong Acid with Strong Base
For example, HCl + NaOH → NaCl + H2O
- Calculate moles of H+ from acid and moles of OH- from base.
- These neutralize, so take the difference to see how much excess remains
- Determine molarity of remaining excess
- Plug into pH = -log[H+] or pOH = -log[OH-]=14-pH
Weak Acid or Base Solution (aka Buffered Solution)
For example, HF or NH3
- Forget about the salt ion if there is any (Cl- or Na+, etc.)
- Write your equation for how the weak acid or base reacts with H2O
Weak acid: HA + H2O ← → A- + H+ (HF + H20 ← → F- + H+)
Weak base: B + H2O ← → BH+ + OH- (NH3 + H20 ← → NH4+ + OH-) - Write an ICE table. For example:
- If you start with just HA (acid) or B (base), plug into:
Ka = [H+]²/[HA]
Kb = [OH-]²/[B] - Or if you’re given A- (conj base) or HB (conj acid), use formula:
[H+] = Ka[acid]/[conj base]
[OH-] = Kb[base]/[conj acid]
Weak Acid or Base with Conjugate Acid or Base (ie mixing a buffered solution)
For example, CH3COOH + CH3COO-
- Use Henderson-Hasselbach:
- If molarity of acid and its conjugate base (or base and its conjugate acid) are the same, then the pH will equal the pKa
Weak Acid with Weak Base
- As long as Ka and Kb values are substantially different, only worry about the stronger one (ie bigger Ka or Kb)
- Plug the larger Ka or Kb into:
[H+] = Ka[acid]/[conj base]
[OH-] = Kb[base]/[conj acid]
Strong Acid or Base with Buffered Solution (ie Weak Acid or Base)
Part 1: Reaction with buffered solution
- Write equation for how the weak acid or base reacts with H2O. Again forget about the salt ion.
Weak acid: HA + H2O ← → A- + H+
Weak base: B + H2O ← → BH+ + OH- - Assume all the H+ or OH- from the strong acid or base will react according to this equation above.
- Determine the number of moles of each compound after all the H+ or OH- is used. If H+ or OH- is not in excess, proceed to Part 2. If it is in excess, proceed to part 3.
Part 2: Reaction of buffered solution when H+ or OH- is not in excess
- Draw ICE chart and assume 0 for [H+] or [OH-] to begin
- Add x to H+ or OH- (and add and subtract accordingly)
- Use Ka or Kb equations to determine H+ or OH- concentration
[H+] = Ka[acid]/[conj base]
[OH-] = Kb[base]/[conj acid]
Part 3: When H+ or OH- is in excess
- Determine how many moles of acid or base are used to neutralize
- Calculate molarity of remaining moles of H+ or OH-
- Calculate accordingly:
-log [H+]= pH
14 + log [OH-] = pH
Strong Acid with Strong Acid or Strong Base with Strong Base (uncommon)
For example, HCl + HI
- Combine total moles of H+ for each acid or OH- for each base.
- Find molarity or H+ or OH- and determine pH
BUFFERED SOLUTIONS
A buffered solution resists pH changes when OH- or protons are added
- Contains either a weak acid and its salt or a weak base and its salt
- These weak acids and bases do not react to completion, but reach equilibrium with their conjugate acids and bases
- As a result, a buffered solution has basic and acidic compounds at once that can react with stronger acids or bases
- Thus, a buffered solution can neutralize H+ or OH- ions to avoid pH change
Buffered Solution Definitions
- Titration Curve: Graph plotting the pH vs mL of a solution as a titrant (reactant) is added
- Half-equivalence Point: the point in a titration when the pH of the solution equals the pKa of the acid (the point halfway between the origin and the equivalence point on both axes)
- Equivalence Point: The point in a titration when enough titrant has been added to react exactly with the substance in solution being titrated.
- Buffering region: The range on a titration curve when the buffered solution neutralizes additional acids or bases, so that the pH barely changes.
- Buffer Capacity: The quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.
DETERMINING THE pH OF SALT SOLUTIONS
Goal: To determine acidity, we want to determinate the concentration of H3O+ or OH- produced from weak acid and base reactions.
- Determine which ion forms with H2O to produce a weak acid or base
- The weaker an acid, the lower the Ka.
- The weaker a base, the lower the Kb.
- Cheat: Find which ion matches your Ka/Kb table of weak acids/bases
2. Find the Ka or Kb for this acid or base in your table.
3. Plug into this equation:
- If you have Kb for weak base, find Ka (equilibrium constant for the H3O+ produced alongside this base).
- If you have Ka for a weak acid, find Kb (equilibrium constant for the OH- produced alongside this acid, i.e. OH- + HCH3COO above).
4. Set the Ka or Kb value you found equal to x²/M
- Long story: the x is [H3O+] or [OH-] derived from ICE worksheets
- Short story: M is the molarity of the acid/base (usually the same as given molarity of salt)
5. Solve for x.
- If you found Ka, x = [H3O+]. So -log (x) = pH
- If you found Kb, x = [OH-]. So 14 + log (x) = pH
FORMULAS
