Strengths of Acids and Bases

Strengths of Acids and Bases

Acid and Base Ionization

• Monoprotic acids: acids with 1 ionizable $H$.
• Polyprotic acids: acids with 2 or more ionizable $H$ (diprotic: 2 $H$, triprotic: 3 $H$).
• Acids lose 1 $H^+$ at a time.
• Ions form in 2 ways:
  • Dissociation: Ionic substances dissolve in water and separate into ions (physical change).
  • Ionization: Covalent substances react with water to form ions (chemical change).
• Example: $HF(aq) + H<em>2O(l) \leftrightarrow F^-(aq) + H</em>3O^+(aq)$

Concentration vs. Strength

• Concentration: # moles dissolved in solution; does NOT equal strength.
• Strength (strong or weak): Extent of ionization or dissociation of an acid or base.
  • How many particles dissociate into ions, NOT the concentration.
• Strong Acid: Ionizes completely in water.
  • Examples: $HCl$, $HBr$, $HI$, $HNO<em>3$, $H</em>2SO<em>4$, $HClO</em>4$
  • Strong electrolyte.
• Weak Acid: Partially ionizes in water.
  • Examples: $HCN$, $HF$, $HC<em>2H</em>3O_2$
  • Aqueous solution contains $H_3O^+$ ions, anions (conjugate base), and non-ionized acid molecules.
  • Weak electrolyte.
• Strong Base: Completely dissociates in water; strong electrolyte.
  • Examples: Hydroxides of groups 1 & 2 (except Be) - $NaOH$, $Ca(OH)<em>2$. Note: $Ca(OH)</em>2$ and $Mg(OH)_2$ are not very soluble.
• Weak Base: Partially ionizes in water; weak electrolyte.
  • Examples: $NH<em>3$, aniline $C</em>6H<em>5NH</em>2$, methylamine $CH<em>3NH</em>2$

Ionization Reaction Extent

• Depends on relative strengths of acids and bases involved
• Strong acid: forward reaction dominates
  • Produces a weak conjugate base which has a low attraction for $H^+$ so reverse reaction does not dominate
  • Ex $HCl + H<em>2O \rightarrow H</em>3O^+ + Cl^-$
  • The Conjugate Base ($Cl^-$) is a weaker base than water so water wins competition over $Cl^-$ for $H^+$
• Weak acid: reverse reaction dominates
  • Most of acid originally placed in solution is still present at equilibrium
  • It has a stronger conjugate base which is better at attracting $H^+$ than water, so as the Conjugate Base forms it combines with the $H^+$ to form the undissociated form of the acid again.
  • Ex $HCOOH + H<em>2O \leftrightarrow H</em>3O^+ + COOH^-$
  • Formic acid is a weak acid, and its Conjugate Base is strong
• Strong base has a weak conjugate acid, and a weak base has a strong conjugate acid.

Acid Dissociation Constant ($K_a$)

• For a weak acid in aqueous solution that doesn't ionize completely: $HA \leftrightarrow H^+ + A^-$
  • Where $HA$ = acid, $H^+$ = hydrogen ion, $A^-$ = conjugate base.
• $K_a$: Ratio of ionized to non-ionized acid concentrations at equilibrium, a quantitative measure of degree of ionization.
• Formula: $K_a = \frac{[H^+][A^-]}{[HA]}$
• Example: $HCN + H<em>2O \leftrightarrow CN^- + H</em>3O^+$ so $K<em>a = \frac{([H</em>3O^+][CN^-])}{[HCN]}$
• $K_a$ value for $HCN$ is $6.2 \times 10^{-10}$.

$K_a$ and Acid Strength

1. Larger $K_a$: Stronger acid - ionizes more completely, higher concentration of ions. A stronger acid will have a higher $[H^+]$ and thus a lower pH
2. Smaller $K_a$: Weaker acid – ionizes less completely, lower concentration of ions and higher concentration of un-ionized acid, it will have a lower $[H^+]$ and higher pH
3. $K_a$ for weak acids are generally known to an accuracy of only $\pm 5\%$, this is called the 5% rule
   • so $[H^+]$ must be less than or equal to 5% of $[HA]$ at equilibrium

Polyprotic Acids and Ionization

• # H does not indicate strength: diprotic and triprotic acids lose $H^+$ one at a time, each successive $H^+$ is more difficult to lose
• All stages of ionization occur in solution, with highest concentration of ions from 1st ionization
• Ex.
  • $H<em>2SO</em>4 + H<em>2O \rightarrow H</em>3O^+ + HSO<em>4^-$ ($H</em>2SO_4$ is a strong acid)
  • $HSO<em>4^- + H</em>2O \rightarrow H<em>3O^+ + SO</em>4^{2-}$ ($HSO_4^-$ is a weak acid)

Base Dissociation Constant ($K_b$)

• Weak base reacts with water to form hydroxide ions and the conjugate acid.
  • $B + H_2O \leftrightarrow HB^+ + OH^-$ (where $B$ = base, $HB^+$ = Conjugate acid)
• $K_b = \frac{[conjugate acid][OH^-]}{[base]} = \frac{[HB^+][OH^-]}{[B]}$
• Example: For
  • $NH<em>3 + H</em>2O \leftrightarrow NH_4^+ + OH^-$
  • $K<em>b = \frac{([NH</em>4^+][OH^-])}{[NH_3]} = 1.8 \times 10^{-5}$
  • $NH_3$ is a weak base
• Measure of degree of dissociation of a base and ability of the base to compete with $OH^-$ for $H^+$
• Strong base has a large $K<em>b$ as more of it is dissociated; weak base has a small $K</em>b$

Solving Equilibrium Problems

• To find $K<em>a$ (or $K</em>b$), substitute molar concentrations for all substances present at equilibrium where rate of forward reaction = rate of reverse reaction
• ICE Chart: Helpful tool.
  • I = Initial Concentration of each species in the reaction mixture
  • C = Change in concentration of each species as the system moves towards equilibrium
  • E = Equilibrium concentration of each species when the system gets to a state of equilibrium

ICE Charts

• Express all concentrations in terms of molarity or appropriate pressure units
• The magnitude of change ($x$) is determined by the balanced chemical equation.
• Given initial concentrations, calculate the equilibrium concentrations using the quadratic formula, if necessary.
• Use equilibrium concentrations to determine K
• The change in each quantity must agree with the reaction stoichiometry (look at the balanced equation!)
• Clearly define what you mean to be represented by “x” and define all other unknown changes in terms of this change.

Math Example 1

• A 0.450 M solution of a weak monoprotic acid has a $[H^+]$ of $2.78 \times 10^{-5}$ M. What is the value of $K_a$ for this acid?

1. Balanced equation: $HA \leftrightarrow H^+ + A^-$
2. ICE chart.
   • Note: the initial values for $H^+$ and $A^-$ are the same.
   • Also, note: the value for $[HA]$ does not change significantly. For the purposes of our calculations, use the calculator answer (don’t worry about sig figs at this time).
   • Write $K_a$ expression, substitute and solve:
   • $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(2.78 \times 10^{-5})^2}{0.450} = 1.72 \times 10^{-9}$

• The change in $[HA]$ is so very small ([H^+] << [HA]) with this, so we can ignore the changes in $[HA]$ and use the concentrations given in the problem.

Math Example 2

• What is the pH of a solution of 0.50 M solution of acetic acid ($HC<em>2H</em>3O<em>2$) at equilibrium, given that the $K</em>a$ for acetic acid is $5.8 \times 10^{-5}$ ?

1. Write the balanced equilibrium equation for the reaction of acetic acid with water to form $H^+$ and $C<em>2H</em>3O_2^-$ ions.
   • $HC<em>2H</em>3O<em>2 \leftrightarrow H^+ + C</em>2H<em>3O</em>2^-$
2. Using the balanced equation, develop your ICE chart. Since acetic acid is a monoprotic acid (it only produces 1 $H^+$ in solution), let x = moles of $H^+$ formed.
3. If x = moles of $H^+$ formed, x also = moles of $C<em>2H</em>3O<em>2^-$ formed and –x = moles of $HC</em>2H<em>3O</em>2$ that react with water. Notice that the concentrations of the products are 0 at the beginning of the reaction.
4. Even though water is a reactant, it is not included in either the expression or the math because its concentration does not change enough to make any difference.

• Because this is a weak acid, it will only ionize at 5% or less. Therefore, the calculations are a bit more streamlined. We can assume that the value for x (the change) is very small compared to the initial acid concentration. This means that the original acid concentration should be used in the calculations, since the change is irrelevant. This makes the change negligible, and it simplifies the math:
• $[H^+] = \sqrt{(K_a)([HA])}$

1. Write $K_a$ expression, substitute and solve for x:
   • $K<em>a = \frac{[H^+][C</em>2H<em>3O</em>2^-]}{[HC<em>2H</em>3O_2]} = \frac{x^2}{0.50} = 5.8 \times 10^{-5}$, $x = 5.4 \times 10^{-3}$
   • Therefore, $pH = -log[H^+] = -log (5.4 \times 10^{-3} M) = 2.27$

Tips for $K<em>a$ and $K</em>b$ Problems

• Always begin with a balanced equation.
• Develop your ICE chart, define x and write your $K_a$ expression.
• Since you are dealing with a weak acid, it will ionize at 5% or less.
• This means that the difference between the concentration of the unionized acid and the concentration of the hydrogen (or hydronium) ions is statistically negligible and can be ignored.
• This makes the math much easier, since you will avoid needing to use the quadratic equation to solve for these concentrations. Remember that calculators can solve quadratic equations. Just learn the feature. Do not get bogged down in the math. Know how to set up the math, that is most important. But using a calculator is crucial for saving time. All calculators have different ways to carry out the funtion, such as Finance, statitics, or solve packages. Take the time to learn. Use your own! Take a calculator to the exam that you understand how it works. Trying to learn to use a new calculator while taking the exam is not wise. Plan in advance and you will be less stressed. Plus you should know how to do the math by hand. Test makers like to give problems designed to trip you up. Do not fall for their games. Keep thinking about whether the answer makes sense or not. Look for the games!} Remember learning to use your calculator can pay big dividends on the final exam.