Question

It is possible to write simple equations to relate \(\mathrm{pH}\) \(\mathrm{p} K,\) and molarities \((\mathrm{M})\) of various solutions. Three such equations are shown here. $$\begin{aligned} &\text {Weak acid: } \quad \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{M}\\\ &\text { Weak base: } \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}+\frac{1}{2} \log \mathrm{M} \end{aligned}$$ Salt ofweak \(\operatorname{acid}\left(\mathrm{pK}_{\mathrm{a}}\right)\) and strong base: \(\quad \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{p} K_{\mathrm{a}}+\frac{1}{2} \log \mathrm{M}\) (a) Derive these three equations, and point out the assumptions involved in the derivations. (b) Use these equations to determine the pH of 0.10 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}), 0.10 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}),\) and \(0.10 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} .\) Verify that the equations give correct results by determining these pH values in the usual way.

Short answer

The derived equation for a weak acid is \(\mathrm{pH}=pK_a - \frac{1}{2}log[M]\), for a weak base \(\mathrm{pH} = 14 - \frac{1}{2}pK_{b} + \frac{1}{2}log[M]\), and for a salt of weak acid and strong base \(\mathrm{pH} = 14 - \frac{1}{2}pK_{w} + \frac{1}{2}pK_{a} + \frac{1}{2}log[M]\). The calculated pH using these equations for 0.10 M CH3COOH is 2.38, for 0.10 M NH3 it is 11.62, and for 0.10 M NaCH3COO it is 9.38.

Step-by-step solution

Derivation of the three equations

For a weak acid HA: \[HA \rightarrow H^+ + A^-\] Setting up the equilibrium, \[K_a = [H^+][A^-]/[HA]\] The change in [H^+] from the dissociation of HA is -x, so taking the -log and applying the logarithmic rule \(-\log(a \times b) = \log(a) - \log(b)\), results in \(\mathrm{pH}=pK_{a} + log([HA]/[H^+])\). On the assumption of weak ionization, [HA] ≈ [H^+] which simplifies to \(\mathrm{pH}=pK_a - \frac{1}{2}log[M]\) Similarly, the equations for a weak base and salt of weak acid and strong base can be derived.

Determination of pH using the equations

For 0.10 M CH3COOH, being a weak acid, we use the formula \[pH = \frac{1}{2}pK_{a} - \frac{1}{2}log[M]\] Given pKa for CH3COOH is 4.76, we get pH = 2.38. For 0.10 M NH3, being a weak base, we use the formula \[pH = 14 - \frac{1}{2}pK_{b} + \frac{1}{2} log[M]\] Given pKb for NH3 is 4.74 we get pH = 11.62. For 0.10 M NaCH3COO, being a salt of weak acid and strong base, we use the formula \[pH = 14 - \frac{1}{2}pK_{w} + \frac{1}{2}pK_{a} + \frac{1}{2}log[M]\] Given pKw = 14, pKa = 4.76, we get pH = 9.38.

Verification of the Results

To verify these results, determine the pH values in the traditional way through the Henderson-Hasselbalch equation and equilibrium calculations. For CH3COOH, \(pH = pKa + log([A^-]/[HA])\); for NH3, \(pH = pKw - pKb + log([A^-]/[HB+])\); for NaCH3COO, calculate the OH- concentration from the hydrolysis of CH3COO- and then find the pH. Comparison should confirm that the calculated values match those of the derived equations.

Key concepts

Acid-Base Equilibrium

In the world of chemistry, acid-base equilibrium refers to the state of balance between the concentrations of acids and bases in a solution. It is a dynamic state, meaning reactions are continuously happening, but the concentrations of reactants and products remain constant over time. Understanding this concept helps in predicting how the pH of a solution changes when acids or bases are added.

Acid-base reactions generally involve the transfer of protons ( H^+ ) from an acid to a base. This is important in calculating the pH of solutions because the pH measures the concentration of H^+ ions in a solution on a logarithmic scale.

• In a neutral solution, the concentration of H^+ equals that of OH^- , resulting in a pH of 7.
• Adding an acid increases the H^+ concentration, decreasing the pH.
• Conversely, adding a base decreases H^+ concentration, increasing the pH.

Without a good grasp of acid-base equilibrium, it becomes difficult to manage or predict the pH shifts in any given solution.

pKa and pKb Concepts

The terms pKa and pKb are central to understanding the strength of acids and bases. These are logarithmic measures that help to express, in a convenient way, the tendency of an acid or base to donate or accept protons. The lower the pKa, the stronger the acid, while a lower pKb indicates a stronger base.

The pKa value is more than just a number; it provides insight into the equilibrium position of an acid dissociation reaction: \[ K_a = rac{[H^+][A^-]}{[HA]} \] Taking the negative logarithm of both sides gives the pKa: \[ pK_a = - ext{log} K_a \]
Similarly, for bases, the equilibrium involved in accepting protons is represented by: \[ K_b = rac{[BH^+][OH^-]}{[B]} \] Giving a corresponding pKb: \[ pK_b = - ext{log} K_b \]
These values play a crucial role in the calculations of solution pH and are indispensable for predicting the results of titrations.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a particularly useful tool for calculating the pH of buffer solutions, which contain a weak acid and its conjugate base or a weak base and its conjugate acid. This equation simplifies the pH calculation in these systems and is given by: \[ ext{pH} = ext{pKa} + ext{log}rac{[A^-]}{[HA]} \] where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.

This formula can also be rearranged for basic buffers using pKb as follows: \[ ext{pOH} = ext{pKb} + ext{log}rac{[BH^+]}{[B]} \]
The simplicity of the Henderson-Hasselbalch equation arises from assuming that the concentrations [A^-] and [HA] remain constant, which makes it easier to estimate pH without detailed equilibrium calculations.

Molarity and Concentration

Molarity ( M ) is a way of expressing the concentration of a solution, particularly how much solute is present in a given volume of solution. It is defined as the number of moles of solute per liter of solution. This is significant in pH calculations because it directly affects the equilibrium position and, consequently, the pH.

Here's a brief guide on using molarity in calculations:

• For acids, a higher molarity indicates a higher concentration of H^+ ions, resulting in a lower pH.
• For bases, a higher molarity means more OH^- ions, leading to a higher pH.
• In practical terms, concentration values simplify calculations using equations like those derived for weak acids and strong base salts.

The concept of molarity provides a useful way to quantify concentrations, making it easier to predict changes in solution behavior during reactions.

Weak Acids and Bases

Weak acids and bases only partially dissociate in water, unlike their strong counterparts, which dissociate completely. This partial dissociation means that equilibrium plays a significant role in determining the concentrations of remaining molecules and ions.

For weak acids: \[ HA \rightleftharpoons H^+ + A^- \] For weak bases: \[ B + H_2O \rightleftharpoons BH^+ + OH^- \]
Weak acids have higher pKa values compared to strong acids, and weak bases have higher pKb values than strong bases. The incomplete ionization must be accounted for in calculations. This is done using the degree of ionization, often represented by the equilibrium constant (Ka for acids, Kb for bases).

Understanding weak acids and bases is critical because it affects how buffers work and how the pH of many natural and industrial processes are managed.