Question

Calculate the pH of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 M\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

Short answer

The pH values for the four solutions are as follows: a. Propanoic Acid Solution: \(pH = 2.94\) b. Sodium Propanoate Solution: \(pH = 8.99\) c. Pure Water: \(pH = 7.00\) d. Mixture Solution: \(pH = 4.87\)

Step-by-step solution

Calculate the pH of Propanoic Acid Solution

To determine the pH for \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\), we can first write down the acid dissociation equation: \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} \leftrightharpoons \mathrm{H}^{+}+\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}\) Let the concentration of dissociated ions be \(x\). We can write the equilibrium concentrations in terms of the initial concentration: \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}: 0.100-x\) \(\mathrm{H}^{+}: x\) \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}: x\) Now, we can write the \(K_\mathrm{a}\) expression: \(K_\mathrm{a}=\dfrac{[\mathrm{H}^{+}][\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}]}{[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}]}=\dfrac{x^2}{0.100-x}\) We can approximate \(0.100-x \approx 0.100\), because \(K_\mathrm{a}\) is very small (weak acid). Therefore, \(x^2 = K_\mathrm{a} \times 0.100\) \(x = \sqrt{1.3 \times 10^{-5} \times 0.100}=1.14×10^{-3}\) Finally, we calculate the pH: \(pH=-\log[\mathrm{H}^{+}]=- \log(1.14×10^{-3})=2.94\)

Calculate the pH of Sodium Propanoate Solution

Sodium propanoate is a salt of a weak acid and a strong base, which will undergo hydrolysis and cause an increase in pH. Since the sodium ion is a spectator ion, we focus on the equilibrium of the propanoate ion, \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}\), in the water, which contributes \(\mathrm{OH}^-\) ions: \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-} + \mathrm{H}_{2} \mathrm{O} \leftrightharpoons \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} + \mathrm{OH}^{-}\) We can find the \(K_\mathrm{b}\) value for the propanoate ion by using the relationship between \(K_\mathrm{a}\), \(K_\mathrm{b}\) and \(K_\mathrm{w}\): \(K_\mathrm{b} = \dfrac{K_\mathrm{w}}{K_\mathrm{a}}\) Let \(x\) be the concentration of propanoate ions that underwent hydrolysis forming hydroxide ions. At equilibrium, the concentrations are: \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}: 0.100-x \) \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}: x\) \(\mathrm{OH}^{-}: x \) Now, we can write the \(K_\mathrm{b}\) expression: \(K_\mathrm{b}=\dfrac{[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}][\mathrm{OH}^{-}]}{[\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}]}\) Using the equilibrium concentrations and calculating \(K_\mathrm{b}\) from \(K_\mathrm{a}\) and \(K_\mathrm{w}\), we can solve for \(x\) to get the hydroxide ion concentration. Then, we can calculate the \(pOH\) and \(pH\) values accordingly.

Calculate the pH of Pure Water

Pure water acts as both an acid and a base and undergoes auto-ionization, resulting in very low concentrations of both hydroxide and hydronium ions. \(\mathrm{H}_{2} \mathrm{O} \leftrightharpoons \mathrm{H}^{+} + \mathrm{OH}^{-}\) At 25 °C, the \(K_\mathrm{w}\) value is \(1.0 \times 10^{-14}\). Since in pure water the concentrations of hydronium and hydroxide ions are equal, we can write: \(K_\mathrm{w}=[\mathrm{H}^{+}]^2\) \(\mathrm{pH}=-\log\left( [\mathrm{H}^{+}] \right)\) We can substitute the \(K_\mathrm{w}\) value into the equations to find the pH of pure water, which should be close to 7.

Calculate the pH of the Mixture Solution

We now need to analyze a mixture containing both the weak acid and its conjugate base, \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\). This is a buffer system. The buffer system's pH can be found using the Henderson-Hasselbalch equation: \(pH = pK_\mathrm{a} + \log \left(\dfrac{[\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^-]}{[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}]}\right)\) We are given the initial concentrations of both species, so we can simply substitute them into the equation to compute the mixture's pH.

Key concepts

Acid-Base Chemistry

Acid-base chemistry is a fundamental part of chemistry that deals with the concepts of acids and bases. It's crucial to understanding how chemical reactions occur in solutions. Acids are substances that donate protons (\(\mathrm{H}^{+}\) ions) when dissolved in water, while bases are substances that accept protons. They can also produce hydroxide ions (\(\mathrm{OH}^{-}\)) in solutions. The pH scale, which ranges from 0 to 14, measures the acidity or basicity of a solution. A pH of less than 7 indicates an acidic solution, a pH of 7 indicates a neutral solution (like pure water), and a pH greater than 7 indicates a basic solution. Understanding these relationships is key when predicting the behavior of chemical species in different environments in acid-base chemistry.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential tool used in chemistry to estimate the pH of buffer solutions, particularly in acid-base chemistry. This equation provides a quick way to calculate the pH of a buffer solution based on the concentrations of the acid and its conjugate base. The equation is written as:\[ pH = pK_a + \log \left(\dfrac{[A^-]}{[HA]}\right) \]where \(pH\) is the potential of hydrogen, \(pK_a\) represents the acid dissociation constant on a logarithmic scale, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid. Utilizing this equation allows chemists to easily determine the pH of a buffer system by knowing the concentrations of its components, making the Henderson-Hasselbalch equation an indispensable tool in laboratory settings and real-world applications.

Buffer Systems

Buffer systems are crucial in maintaining a stable pH environment in biological systems and industrial processes. A buffer consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. These systems resist changes in pH when small amounts of acid or base are added. In a buffer solution, the weak acid neutralizes added bases, and the conjugate base neutralizes added acids. This buffering action is crucial for many biochemical processes in organisms. For example, human blood maintains a very narrow pH range due to the presence of carbonate-based buffers. Buffers are also widely employed in laboratory reagents and industrial processes, where the control of pH is essential to the success of the chemical reactions being performed.

Acid Dissociation Constant (Ka)

The acid dissociation constant, abbreviated as \(K_a\), is a quantitative measure of the strength of an acid in a solution. It reflects how easily an acid can donate protons to water. A large \(K_a\) value indicates a strong acid that dissociates completely, while a smaller \(K_a\) characterizes a weak acid with less dissociation in water.Mathematically, the \(K_a\) is expressed as:\[ K_a = \dfrac{[H^+][A^-]}{[HA]} \]where \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the undissociated acid. In calculations, weak acids are often associated with smaller \(K_a\) values, which allows assumptions such as ignoring the dissociated amount when it's much smaller than the initial concentration. Understanding \(K_a\) is fundamental for interpreting chemical reactions and predicting the behavior of different acid-base systems.