Question

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

Short answer

The pH values for each solution are: a. Propanoic acid: pH ≈ 2.94 b. Sodium propanoate: pH ≈ 8.44 c. Pure water: pH = 7.00 d. Mixture of propanoic acid and sodium propanoate: pH ≈ 4.89

Step-by-step solution

Calculate the pH of the propanoic acid solution

Given a solution of 0.100 M propanoic acid (HC3H5O2) and its dissociation constant, Ka = 1.3 × 10^(-5). This is a weak acid, and we can set up an equilibrium expression to find its ionization degree: \[HC_3H_5O_2 \rightleftharpoons H^+ + C_3H_5O_2^-\] To find the ionization degree (x), we can write the expression for Ka: \[Ka = \frac{[H^+][C_3H_5O_2^-]}{[HC_3H_5O_2]}\] Since the initial concentration of propanoic acid is 0.100 M and assuming the ionization degree is small, we can write: \[Ka = \frac{x^2}{0.100 - x}\] Now, we'll solve for x, which is also the [H+] concentration: \[x^2 = Ka(0.100 - x) = 1.3 \times 10^{-5}(0.100)\] \[x = \sqrt{1.3 \times 10^{-6}}\] Since x is very small compared to 0.100, x ≈ 1.14 × 10^(-3) M Now we can calculate the pH: \[pH = -\log[H^+]\] \[pH = -\log(1.14 \times 10^{-3}) ≈ 2.94\]

Calculate the pH of the sodium propanoate solution

Since sodium propanoate (NaC3H5O2) completely dissociates in water, the solution will be 0.100 M in C3H5O2^- ions. We can use the Kb value to determine the pH: \[Kb = \frac{Kw}{Ka}\] Where Kw is the ion product of water (1.0 × 10^(-14)). \[Kb = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}} = 7.69 \times 10^{-10}\] Now, we can write the equilibrium expression of sodium propanoate's conjugate base (C3H5O2^-): \[C_3H_5O_2^- + H_2O \rightleftharpoons HC_3H_5O_2 + OH^-\] Using Kb, we can write the ionization expression as: \[Kb = \frac{[HC_3H_5O_2][OH^-]}{[C_3H_5O_2^-]}\] We have the initial concentration of C3H5O2^- (0.100 M), and we'll assume that the hydroxide concentration is x: \[Kb = \frac{x^2}{0.100 - x} = 7.69 \times 10^{-10}\] \[x = \sqrt{7.69 \times 10^{-11}} ≈ 2.77 \times 10^{-6}\] Now, we can calculate the pOH: \[pOH = -\log[OH^-]\] \[pOH = -\log(2.77 \times 10^{-6}) ≈ 5.56\] Finally, we can find the pH: \[pH = 14 - pOH = 14 - 5.56 ≈ 8.44\]

Calculate the pH of pure water

For pure water, the concentration of H+ and OH- ions is equal and can be obtained using Kw: \[Kw = [H^+][OH^-]\] \[1.0 \times 10^{-14} = [H^+][H^+]\] \[H^+ = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\] The pH of pure water is: \[pH = -\log[H^+] = -\log(1.0 \times 10^{-7}) = 7\]

Calculate the pH of the mixture containing propanoic acid and sodium propanoate

For this step, we can use the Henderson-Hasselbalch equation: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] \[pH = -\log(1.3 \times 10^{-5}) + \log \frac{0.100}{0.100}\] \[pH = 4.89 + \log(1) = 4.89\] In conclusion, the pH values for each solution are: a. Propanoic acid: pH = 2.94 b. Sodium propanoate: pH = 8.44 c. Pure water: pH = 7.00 d. Mixture of propanoic acid and sodium propanoate: pH = 4.89

Key concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is essential to predict how acids and bases behave in solutions. When we talk about equilibrium in this context, we're looking at the balance between the forms of an acid or base in solution. For example, propanoic acid (\(HC_3H_5O_2\)) is a weak acid.
The equilibrium involves the reversible reaction:

• \(HC_3H_5O_2 \rightleftharpoons H^+ + C_3H_5O_2^-\)

This shows the dissociation of propanoic acid into hydrogen ions (\(H^+\)) and propanoate ions (\(C_3H_5O_2^-\)).
In equilibrium, the rate at which \(HC_3H_5O_2\) dissociates equals the rate at which \(H^+\) and \(C_3H_5O_2^-\) recombine.
By knowing the equilibrium constant (\(K_a\)), you can determine the extent of this dissociation.
This is done by setting up the ionization expression, using the definition of \(K_a\):

• \(K_a = \frac{[H^+][C_3H_5O_2^-]}{[HC_3H_5O_2]}\)

This helps predict the pH, a measure of the acidity of the solution.
When calculating pH, assumptions might be needed, such as ignoring small changes in concentration when they are negligible compared to initial concentration.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is very useful when dealing with buffer solutions. A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid, able to maintain a stable pH.
This equation helps estimate the pH of such a solution using its components' concentrations:

• \[ pH = pK_a + \log \frac{[A^-]}{[HA]} \]

Here, \(pK_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)), \([A^-]\) is the concentration of the base form, and \([HA]\) is the concentration of the acid form.
In our example, turning to the mixture of propanoic acid (\(HC_3H_5O_2\)) and sodium propanoate (\(NaC_3H_5O_2\)) shows a classic buffer.
Using equal concentrations of acid and conjugate base simplifies the calculations, since the log of 1 equals 0:
\(pH = pK_a\). This is why the pH of the mixture is close to the \(pK_a\) of propanoic acid.
The Henderson-Hasselbalch equation is particularly handy because it avoids solving quadratic equations typical in buffer calculations.

Ionization Constant

The ionization constant is a critical parameter in understanding acid and base strengths. For acids, this constant is often represented by \(K_a\) and for bases by \(K_b\).
It provides information on how much an acid or base dissociates in solution. A larger \(K_a\) or \(K_b\) value indicates a stronger acid or base, which dissociates more in solution.
For acids, the ionization constant \(K_a\) is used in equilibria like:

• \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

In this formula, \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the undissociated acid.
The ionization constants are also pivotal in calculating pH or pOH of the solutions, ensuring accurate predictions of the acidic or basic nature.
For example, knowing \(K_a\) of propanoic acid helps determine its pH when it dissociates in water.
This enables us to apply these constants in conjunction with concentration changes and assumptions in equilibrium expressions and calculations.