Reactions between strong acids and strong bases

When a strong acid and a strong base are mixed, they undergo a complete and quantitative reaction, meaning the reaction goes to completion with no significant reverse reaction. Strong acids and strong bases are fully ionized in water, so their interaction is straightforward and predictable.

The reaction can be represented by a net ionic equation

${\text{H}}^{+}(\text{aq})+{\text{OH}}^{-}(\text{aq})\to {\text{H}}_2\text{O}(\text{l})$

Key characteristics:

• Complete ionization - Strong acids donate all their H⁺ ions, and strong bases provide all their OH^- ions in solution.
• Neutralization - The H⁺ ions from the acid combine with OH^- ions from the base to form water, resulting in a neutral product.
• Excess reagent determines pH - If one reactant is in excess, the pH of the resulting solution depends on the concentration of the leftover H⁺ or OH^- ions.

This reaction is the simplest form of neutralization, often resulting in a solution with a pH close to 7 if the reactants are in equal molar amounts.

Reactions between weak acids and strong bases

Mixing a weak acid with a strong base results in a reaction that also proceeds quantitatively, but the behavior of the weak acid introduces additional complexity. A weak acid only partially ionizes in water, meaning not all of its molecules release H⁺ ions.

The general reaction is represented as

$\text{HA}(\text{aq})+{\text{OH}}^{-}(\text{aq})\rightleftharpoons {\text{A}}^{-}(\text{aq})+{\text{H}}_2\text{O}(\text{l})$

Key characteristics

• Partial ionization of weak acid - Only a small fraction of HA dissociates into H⁺ and A^-, but the strong base fully provides OH^- ions.
• Formation of conjugate base - The OH^- ions react with HA to form water and the conjugate base A^-, shifting the equilibrium to the right.

pH determination based on excess

The pH of the resulting solution depends on the relative amounts of reactants:

• If the weak acid is in excess, a buffer solution forms (a mixture of HA and A^-), and pH is calculated using the Henderson-Hasselbalch equation.
• If the strong base is in excess, the pH is determined by the concentration of excess OH^- ions.
• If equimolar amounts are mixed, the solution is slightly basic due to the hydrolysis of A^-, represented by:

${\text{A}}^{-}(\text{aq})+{\text{H}}_2\text{O}(\text{l})\rightleftharpoons \text{HA}(\text{aq})+{\text{OH}}^{-}(\text{aq})$

This reaction highlights the importance of considering the extent of ionization when predicting solution properties.

Reactions between weak bases and strong acids

When a weak base reacts with a strong acid, the reaction is also quantitative. A weak base only partially accepts protons (H⁺ ions) in water, while a strong acid is fully ionized, donating all its H⁺ ions.

The reaction can be represented as

$\text{B}(\text{aq})+{\text{H}}_3{\text{O}}^{+}(\text{aq})\rightleftharpoons {\text{HB}}^{+}(\text{aq})+{\text{H}}_2\text{O}(\text{l})$

Key characteristics

• Partial protonation of weak base - The weak base B does not fully accept protons, but the strong acid provides H⁺ (as H₃O⁺) completely.
• Formation of conjugate acid - The reaction forms the conjugate acid HB⁺ and water, shifting the equilibrium forward.

The resulting pH depends on the relative quantities

• If the weak base is in excess, a buffer solution forms (a mixture of B and HB⁺), and pH is determined using the Henderson-Hasselbalch equation.
• If the strong acid is in excess, the pH is calculated from the concentration of excess H₃O⁺ ions.
• If equimolar amounts are mixed, the solution is slightly acidic due to the hydrolysis of HB⁺, represented by:

${\text{HB}}^{+}(\text{aq})+{\text{H}}_2\text{O}(\text{l})\rightleftharpoons \text{B}(\text{aq})+{\text{H}}_3{\text{O}}^{+}(\text{aq})$

Understanding the behavior of weak bases in these reactions is crucial for predicting the nature of the resulting solution.

Reactions between weak acids and weak bases

When a weak acid and a weak base are mixed, the reaction does not go to completion but instead reaches an equilibrium state. Both reactants only partially ionize, so the products and reactants coexist in significant amounts.

The reaction is represented as

$\text{HA}(\text{aq})+\text{B}(\text{aq})\rightleftharpoons {\text{A}}^{-}(\text{aq})+{\text{HB}}^{+}(\text{aq})$

Key characteristics

• Partial reaction - Since both HA and B are weak, only a fraction reacts to form the conjugate base A^- and conjugate acid HB⁺.
• Equilibrium mixture - The solution contains a mixture of reactants (HA and B) and products (A^- and HB⁺), and the extent of reaction depends on their relative strengths.
• pH determination - Calculating the pH is more complex and often requires considering the equilibrium constants (K_a for the acid and K_b for the base) of the species involved.

This type of reaction illustrates the nuanced behavior of weak species in solution, where neither dominates completely.

Determining pH in various acid-base mixtures

The pH of a solution resulting from an acid-base reaction depends on the nature and relative amounts of the reactants. Knowing how to assess the major species present after the reaction is key to predicting whether the solution will be acidic, basic, or neutral.

Strong acid and strong base

• Excess strong acid - pH is determined by the concentration of remaining H⁺ ions.
• Excess strong base - pH is determined by the concentration of remaining OH^- ions.
• Equimolar amounts - pH is approximately 7 (neutral) since the reaction forms only water.

Weak acid and strong base

• Excess weak acid - Forms a buffer; pH is calculated using the Henderson-Hasselbalch equation.
• Excess strong base - pH is based on excess OH^- concentration.
• Equimolar amounts - Slightly basic due to hydrolysis of the conjugate base A^-; pH is determined from the equilibrium.

Weak base and strong acid

• Excess weak base - Forms a buffer; pH is calculated using the Henderson-Hasselbalch equation.
• Excess strong acid - pH is based on excess H₃O⁺ concentration.
• Equimolar amounts - Slightly acidic due to hydrolysis of the conjugate acid HB⁺; pH is determined from the equilibrium.

Weak acid and weak base

pH depends on the relative strengths of the acid and base, calculated using equilibrium constants.

Using the Henderson-Hasselbalch equation for buffer solutions

A buffer solution resists changes in pH when small amounts of acid or base are added. Buffers form in mixtures where a weak acid and its conjugate base (or a weak base and its conjugate acid) are both present, often occurring when a weak acid reacts with a strong base or a weak base reacts with a strong acid.

Henderson-Hasselbalch equation

$\text{pH}={\text{pK}}_{\text{a}}+\log \left(\frac{[{\text{A}}^{-}]}{[\text{HA}]}\right)$

Where:

• pH = The negative logarithm of the hydrogen ion concentration
• pK_a = The negative logarithm of the acid dissociation constant (K_a) for the weak acid HA
• [A^-] = Concentration of the conjugate base
• [HA] = Concentration of the weak acid

For a weak base and its conjugate acid, the equation can be adapted using pK_b and the concentrations of B and HB⁺, though typically it is expressed in terms of the conjugate acid's pK_a.

Application in buffer systems

• Calculating pH of buffers - This equation allows precise determination of pH in solutions where a weak acid and its conjugate base coexist.
• Predicting buffer behavior - It helps predict how the pH will change with the addition of small amounts of acid or base, as the ratio of [A^-]/[HA] adjusts to maintain pH stability.

Worked example - Calculating pH of a buffer solution

A buffer solution is prepared by mixing 0.50 moles of acetic acid (HA, with K_a = 1.8 × 10^-5, so pK_a = 4.74) and 0.30 moles of its conjugate base (A^-) in a total volume of 1.0 L. Calculate the pH of this buffer.

Step 1: Identify the formula

$\text{pH}={\text{pK}}_{\text{a}}+\log \left(\frac{[{\text{A}}^{-}]}{[\text{HA}]}\right)$

Step 2: Substitute the values

Concentrations are equal to moles per liter since the total volume is 1.0 L:

• [HA] = 0.50 M
• [A^-] = 0.30 M
• pK_a = 4.74

$\text{pH}=4.74+\log \left(\frac{0.30}{0.50}\right)$

Step 3: Calculate the logarithm

$\log \left(\frac{0.30}{0.50}\right)=\log (0.6)=-0.222$

Step 4: Compute the pH

$\text{pH}=4.74-0.222=4.52$

The pH of the buffer solution is approximately 4.52.

Worked example - Determining pH with excess strong base

A solution is prepared by mixing 0.025 moles of a weak acid HA with 0.035 moles of a strong base NaOH in a total volume of 0.50 L. Calculate the pH of the resulting solution, assuming complete reaction of the strong base.

Step 1: Determine the reaction outcome

$\text{HA}(\text{aq})+{\text{OH}}^{-}(\text{aq})\to {\text{A}}^{-}(\text{aq})+{\text{H}}_2\text{O}(\text{l})$

• Initial moles of HA = 0.025
• Initial moles of OH^- = 0.035
• Moles of OH^- in excess = 0.035 - 0.025 = 0.010 moles

Step 2: Calculate the concentration of excess OH^-

Total volume = 0.50 L

$[{\text{OH}}^{-}]=\frac{0.010}{0.50}=0.020\text{ M}$

Step 3: Calculate pOH

$\text{pOH}=-\log (0.020)=1.70$

Step 4: Convert to pH

$\text{pH}=14.00-\text{pOH}=14.00-1.70=12.30$

The pH of the solution is approximately 12.30, reflecting the excess strong base.