Acid and Bases Notes

Introduction to Acids and Bases

Arrhenius Model (1800s)

• An acid in water produces hydrogen ($H^+$) ions.
• A base in water produces hydroxide ($OH^−$) ions.
• Example:
  • $HCl(g) \rightarrow H^+(aq) + Cl^−(aq)$
  • $NaOH(s) \rightarrow Na^+(aq) + OH^−(aq)$
• Arrhenius received the Nobel Prize for demonstrating the significance of $H^+(aq)$ and $OH^−(aq)$ ions in acid-base chemistry.

Problem with Arrhenius Model

• $H^+$ does not exist freely in aqueous solutions; it associates with water molecules.
• Chemists represent aqueous $H^+$ ions as hydronium ions ($H_3O^+(aq)$).

Brønsted–Lowry Definition

• A Brønsted–Lowry acid is a proton ($H^+$) donor.
• A Brønsted–Lowry base is a proton ($H^+$) acceptor.
• Example:
  • $HCl(g) + H<em>2O(l) \rightarrow H</em>3O^+(aq) + Cl^−(aq)$
  • Here, $HCl$ is a Brønsted–Lowry acid because it donates a proton.
  • $H_2O$ is a Brønsted–Lowry base because it accepts a proton.

Brønsted–Lowry Acids

• A Brønsted–Lowry acid must contain a hydrogen atom.
• Common examples: $HCl$ (hydrochloric acid), $H<em>2SO</em>4$ (sulfuric acid), $HBr$ (hydrobromic acid), $HNO_3$ (nitric acid).

Naming Acids

• Names of acids are derived from the anions formed when they dissolve in water.
• Anions ending in -ide: Add the prefix hydro- and change -ide to -ic acid.
  • Example: $Cl^−$ (chloride) $\rightarrow$ $HCl$ (hydrochloric acid).

Naming Polyatomic Anions

• Polyatomic anions ending in -ate: Change -ate to -ic acid.
  • Example: $SO<em>4^{2−}$ (sulfate) $\rightarrow$ $H</em>2SO_4$ (sulfuric acid).
• Polyatomic anions ending in -ite: Change -ite to -ous acid.
  • Example: $SO<em>3^{2−}$ (sulfite) $\rightarrow$ $H</em>2SO_3$ (sulfurous acid).

Brønsted–Lowry Bases

• A Brønsted–Lowry base is a proton acceptor, requiring it to form a bond with a proton.

Conjugate Acid-Base Pairs

• A conjugate acid-base pair differs by only one proton.
• Example: $HBr$ and $Br^−$ form a conjugate acid-base pair; $H<em>2O$ and $H</em>3O^+$ are another conjugate acid-base pair.
• The conjugate base of $H<em>3PO</em>4$ is $H<em>2PO</em>4^−$, not $PO_4^{3−}$.

Activity: Conjugate Acid-Base Pairs

• Examples:
  • Acid: $H<em>2CO</em>3$, Conjugate Base: $HCO_3^−$
  • Acid: $H<em>3PO</em>4$, Conjugate Base: $H<em>2PO</em>4^−$
  • Acid: $HPO<em>4^{2−}$, Conjugate Base: $PO</em>4^{3−}$
  • Acid: $NH<em>4^+$, Conjugate Base: $NH</em>3$
  • Acid: $H_2O$, Conjugate Base: $OH^−$

Activity 2: Conjugate Acid-Base Pairs

• Examples:
  • Base: $SO<em>4^{2−}$, Conjugate Acid: $HSO</em>4^−$
  • Base: $HCO<em>3^−$, Conjugate Acid: $H</em>2CO_3$
  • Base: $NH<em>2^−$, Conjugate Acid: $NH</em>3$
  • Base: $ClO<em>2^−$, Conjugate Acid: $HClO</em>2$
  • Base: $H<em>2PO</em>4^−$, Conjugate Acid: $H<em>3PO</em>4$

Activity 3: Conjugate Acid-Base Pairs

• $HCO<em>3^− + H</em>3PO<em>4 \rightleftharpoons H</em>2CO<em>3 + H</em>2PO_4^−$
  • Base: $HCO<em>3^−$, Acid: $H</em>3PO<em>4$, Conjugate Acid: $H</em>2CO<em>3$, Conjugate Base: $H</em>2PO_4^−$

Amphoteric Substances

• A substance that can act as either an acid or a base.
• Water is the most common amphoteric substance.
• Bicarbonate ion ($HCO_3^−$) is another example:
  • $HCO<em>3^−(aq) + OH^−(aq) \rightarrow CO</em>3^{2−}(aq) + H_2O(l)$
    • Acid: $HCO<em>3^−$, Base: $OH^−$, Conjugate Base: $CO</em>3^{2−}$, Conjugate Acid: $H_2O$
  • $HCO<em>3^−(aq) + H</em>3O^+(aq) \rightarrow H<em>2CO</em>3(aq) + H_2O(l)$
    • Base: $HCO<em>3^−$, Acid: $H</em>3O^+$, Conjugate Acid: $H<em>2CO</em>3$, Conjugate Base: $H_2O$

Acid and Base Strength

• Strong acids and bases ionize completely.
• Weak acids and bases ionize to a small extent (small fraction of molecules ionize).
• Strong acids dissolve in water, dissociating 100% into ions.
  • $HCl(g) + H<em>2O(l) \rightarrow H</em>3O^+(aq) + Cl^−(aq)$
  • A single reaction arrow indicates complete dissociation.

Strong Acids

• Strong acids are strong electrolytes.

List of Strong Acids

• $HCl$ (hydrochloric acid)
• $HBr$ (hydrobromic acid)
• $HI$ (hydroiodic acid)
• $HNO_3$ (nitric acid)
• $HClO_3$ (chloric acid)
• $HClO_4$ (perchloric acid)
• $H<em>2SO</em>4$ (sulfuric acid) - only one $H^+$ ionizes completely

Weak Acids

• When a weak acid dissolves in water, only a small fraction dissociates into ions.
• Reversible arrows are used to show this.
• Weak acids are weak electrolytes.

Common Weak Acids

• $CH_3COOH$ (acetic acid) - vinegar
• $H<em>2CO</em>3$ (carbonic acid) - soda, blood
• $H<em>3C</em>6H<em>5O</em>7$ (citric acid) - fruit, soda
• $HF$ (hydrofluoric acid) - glass etching, semiconductor manufacturing
• $HOCl$ (hypochlorous acid) - sanitize pools and drinking water
• $HC<em>3H</em>5O_3$ (lactic acid) - milk
• $H<em>4C</em>4O_5$ (malic acid) - fruit
• $H<em>2C</em>2O_4$ (oxalic acid) - nuts, cocoa, parsley
• $H<em>3PO</em>4$ (phosphoric acid) - soda, blood
• $H<em>2C</em>4H<em>4O</em>6$ (tartaric acid) - candy, grapes

Polyprotic Acids

• Acids that contain more than one acidic hydrogen and can donate more than one $H^+$ ion.
• Acids donate one $H^+$ ion at a time in steps.
• Examples: triprotic acids, diprotic acids

Strong Bases

• When a strong base dissolves in water, 100% of it dissociates into ions.
  • $NaOH(s) + H_2O(l) \rightarrow Na^+(aq) + OH^−(aq)$
• Examples: $LiOH$, $NaOH$, $KOH$, $RbOH$, $CsOH$, $Ca(OH)<em>2$, $Sr(OH)</em>2$, $Ba(OH)_2$
• Strong bases are strong electrolytes.

Weak Bases

• When a weak base dissolves in water, only a small fraction dissociates into ions.

Common Weak Bases

• Weak bases are weak electrolytes.
• Examples:
  • $NH_3$ (ammonia) - glass cleaners
  • $CH<em>3NH</em>2$ (methylamine) - herring brine
  • $(CH<em>3)</em>3N$ (trimethylamine) - rotting fish

Dissociation of Water

• Water self-ionizes to a very small extent:
  • $H<em>2O(l) + H</em>2O(l) \rightleftharpoons H_3O^+(aq) + OH^−(aq)$
• The product of $[H_3O^+]$ and $[OH^−]$ is constant at a given temperature:
  • $K<em>w = [H</em>3O^+][OH^−]$
• At 25°C, $K_w = 1.0 × 10^{−14}$; in a neutral solution:
  • $[H_3O^+] = [OH^−] = 1.0 × 10^{−7} M$

Ion-Product Constant of Water, $K_w$

• Neutral solution: $[H_3O^+] = [OH^−]$
• Acidic solution: [H_3O^+] > [OH^−]
• Basic solution: [OH^−] > [H_3O^+]

Neutral, Acidic, and Basic Aqueous Solutions at 25°C

• $K<em>w = [H</em>3O^+][OH^−] = 1.0 × 10^{−14}$ for any aqueous solution at 25°C

Calculating $H_3O^+$ and $OH^−$ Concentrations

• When $[H_3O^+]$ is known:
  • $[OH^−] = \frac{K<em>w}{[H</em>3O^+]} = \frac{1.0 × 10^{−14}}{[H_3O^+]}$
• When $[OH^−]$ is known:
  • $[H<em>3O^+] = \frac{K</em>w}{[OH^−]} = \frac{1.0 × 10^{−14}}{[OH^−]}$

Activity: Calculating $H_3O^+$ and $OH^−$ Ion Concentrations

• Given $[OH^-]$ concentrations, calculate $[H_3O^+]$.
  1. $[OH^-] = 1.0 × 10^{−8} M$, $[H_3O^+] = 1.0 × 10^{−6} M$, Acidic
  2. $[OH^-] = 0.010 M$, $[H_3O^+] = 1.0 × 10^{−12} M$, Basic

Activity: Strong Acid and Base Solutions

• What is the $[H_3O^+]$ and $[OH^−]$ for each solution?
  • 0.10 M $HCl$: $[H_3O^+] = 0.10 M$, $[OH^−] = 1.0 × 10^{−13} M$
  • 0.10 M $NaOH$: $[H_3O^+] = 1.0 × 10^{−13} M$, $[OH^−] = 0.10 M$
  • 0.00010 M $HCl$: $[H_3O^+] = 1.0 × 10^{−4} M$, $[OH^−] = 1.0 × 10^{−10} M$

The pH Scale

• The pH of a solution is the negative logarithm (base 10) of the $H_3O^+$ concentration:
  • $pH = −log[H_3O^+]$
• Examples:
  1. If $[H_3O^+] = 1.0 × 10^{−1} M$, $pH = 1.00$
  2. If $[H_3O^+] = 1.0 × 10^{−5} M$, $pH = 5.00$
  3. If $[H_3O^+] = 1.0 × 10^{−7} M$, $pH = 7.00$
  4. If $[H_3O^+] = 1.0 × 10^{−11} M$, $pH = 11.00$
• Solutions with pH < 7 are acidic; pH > 7 are basic.

Calculating pH

• $pH = −log[H_3O^+]$
• The lower the pH, the higher the concentration of $H_3O^+$.
  • Acidic solution: $pH < 7, [H_3O^+] > 1 × 10^{−7}$
  • Neutral solution: $pH = 7, [H_3O^+] = 1 × 10^{−7}$
  • Basic solution: pH > 7, [H_3O^+] < 1 × 10^{−7}

Calculating pH Using a Calculator

• A logarithm has the same number of decimal places as there are digits in the original number.
• Example: If $[H_3O^+] = 1.2 × 10^{−5} M$, what is its pH?
  • $pH = −log[H_3O^+] = −log(1.2 × 10^{−5}) = −(−4.92) = 4.92$
  • The solution is acidic because pH < 7.

Activity: Calculating pH

• The hydroxide ion concentration in a soil sample was determined to be $5.0 × 10^{−7} M$.
  1. What is the pH of the soil? $pH = 6.30$
  2. Is the soil acidic or basic? Acidic

Activity Solutions: Calculating pH

1. 0. 00085 M HCl
   • HCl is a strong acid. $[H_3O^+] = 0.00085 M$
   • $pH = −log[H_3O^+] = −log(0.00085) = 3.07$
2. 0. 010 M NaOH
   • NaOH is a strong base. $[OH^−] = 0.010 M$
   • $[H<em>3O^+] = \frac{K</em>w}{[OH^−]} = \frac{1.0 × 10^{−14}}{0.010} = 1.0 × 10^{−12} M$
   • $pH = −log[H_3O^+] = −log(1.0 × 10^{−12}) = 12.00$
3. 0 M $HNO_3$
   • $HNO<em>3$ is a strong acid. $[H</em>3O^+] = 1.0 M$
   • $pH = −log[H_3O^+] = −log(1.0) = 0.00$

Calculating $OH^−$ and $H_3O^+$ Concentrations

• Example: A lake has an $[OH^−] = 1.0 × 10^{−9} M$.

• Is the lake considered dead?

  • $[H<em>3O^+] = \frac{K</em>w}{[OH^−]} = \frac{1.0 × 10^{−14}}{1.0 × 10^{−9}} = 1.0 × 10^{−5} M$
  • $pH = −log[H_3O^+] = −log(1.0 × 10^{−5}) = 5.00$

Calculating $[H_3O^+]$ from pH

• $pH = −log[H_3O^+]$

• An alternative form:

  • $[H_3O^+] = 10^{−pH}$

• Example: If $pH = 6.00$, $[H_3O^+] = 10^{−6.00} = 1.0 × 10^{−6} M$

• Follow-up: If the pH of a soil sample is 6.20, what is the $[H_3O^+]$?

  • $[H_3O^+] = 10^{−6.20} = 6.3 × 10^{−7} M$

Calculating the pH of a Base

• What is the pH of a $1.0 × 10^{−3} M$ $KOH$ solution?

• $KOH$ is a strong base, so it dissociates completely.

  $[OH^−] = 1.0 × 10^{−3} M$

  $K<em>w = 1.0 × 10^{−14} = [H</em>3O^+][OH^−]$

  $[H_3O^+] = \frac{1.0 × 10^{−14}}{1.0 × 10^{−3}} = 1.0 × 10^{−11} M$

  $pH = −log[H_3O^+] = −log(1.0 × 10^{−11}) = 11.00$

Calculating $[OH^−]$ from pH

• What is the $[OH^−]$ of a solution with $pH = 5.00$?

  $[H_3O^+] = 10^{−pH} = 10^{−5.00} = 1.0 × 10^{−5}$

  $K<em>w = 1.0 × 10^{−14} = [H</em>3O^+][OH^−]$

  $[OH^−] = \frac{1.0 × 10^{−14}}{1.0 × 10^{−5}} = 1.0 × 10^{−9} M$

Common Acid–Base Reactions

A. Reactions of Acids with Hydroxide Bases

• Neutralization reaction: An acid-base reaction that produces a salt and water.

• $HA(aq) + MOH(aq) \rightarrow H_2O(l) + MA(aq)$

  • Acid + Base $\rightarrow$ Water + Salt

  • The acid ($HA$) donates a proton ($H^+$) to the base ($OH^−$) to form water ($H_2O$).

  • The anion ($A^−$) from the acid combines with the cation ($M^+$) from the base to form the salt ($MA$).

How to Write a Balanced Equation for a Neutralization Reaction Between HA and MOH

• Example: Write a balanced equation for the reaction of $Mg(OH)_2$ with $HCl$.

  [[1]] Identify the acid and base and draw $H_2O$ as one product.

• $HCl(aq) + Mg(OH)<em>2(aq) \rightarrow H</em>2O(l) + Salt$

  [[2]] Balance the equation. Place a 2 to balance $O$ and $H$.

• $Mg(OH)<em>2(aq) + 2HCl(aq) \rightarrow 2H</em>2O(l) + MgCl_2(aq)$

  • Place a 2 to balance $Cl$

Titration

• Determining an unknown molarity from titration data requires three operations:
  • Volume of Base $\rightarrow$ Moles of Base $\rightarrow$ Moles of Acid $\rightarrow$ Volume of Acid

Determining the Volume of a Base Solution from Titration

• Example: How many milliliters (mL) of a 0.610 M $NaOH$ solution are needed to neutralize 20.0 mL of a 0.245 M $H<em>2SO</em>4$ solution?
  • $H<em>2SO</em>4(aq) + 2NaOH(aq) \rightarrow Na<em>2SO</em>4(aq) + 2H_2O(l)$

Titration Calculations

[[1]] Calculate moles of $H_2SO_4$:

• $moles \ of \ H<em>2SO</em>4 = \frac{0.245 \ mol \ H<em>2SO</em>4}{1000 \ mL \ soln} × 20.0 \ mL \ soln = 4.90 × 10^{− three} \ mol \ H<em>2SO</em>4$

  [[2]] From stoichiometry: 1 mol $H<em>2SO</em>4$ = 2 mol $NaOH$.

• $moles \ of \ NaOH = 2 × 4.90 × 10^{− three} = 9.80 × 10^{− three} \ mole$

  [[3]] Calculate volume of $NaOH$:

  $Volume \ of \ NaOH = \frac{9.80 × 10^{− three} \ mol \ NaOH }{\frac{0.610 \ mol}{Liter \ soln}} = 0.0161 \ L = 16.1 \ mL$