Question

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.100 M\) HONH \(_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 M\) HONH \(_{3}\) Cl c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing 0.100 \(M \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

Short answer

a. pH of 0.100 M HONH2: \(pH = 11.15\) b. pH of 0.100 M HONH3Cl: \(pH = 5.02\) c. pH of pure H2O: \(pH = 7\) d. pH of the mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl: \(pH = 8.03\)

Step-by-step solution

a. pH of 0.100 M HONH2 with Kb = 1.1 × 10−8

This is a weak base, so we will use the Kb (base dissociation constant) to establish a reaction for the base in water: \( HONH_{2} + H_{2}O \rightleftharpoons OH^- + HONH_{3} \) Now, let's create the ICE (Initial, Change, Equilibrium) table for this reaction: | Base | -OH | +HONH3 | |---------|-----|--------| | Initial |0.100 M| 0 |0 | | Change |-x | +x | +x | | Equilibrium |0.100-x | x | x | \[ Kb = \dfrac{[OH^-][HONH_{3}]}{[HONH_{2}]} = \dfrac{x^2}{0.100 - x} = 1.1 \times 10^{-8} \] Now we'll need to solve this equation for x. Since Kb is very small, we don't have to consider the effect of the decrease in the initial concentration of HONH2 and can assume x is very small compared to 0.100 M. Therefore, x²/0.100 = 1.1 × 10⁻⁸. To find the pH, let's determine the pOH first: \( pOH = -\log_{10} [OH^-] = -\log_{10}x \) Now, find the pH using the following formula: \( pH = 14 - pOH \)

b. pH of 0.100 M HONH3Cl

This is a salt resulting from a weak base (HONH2) and a strong acid (HCl). When dissolved in water, it undergoes hydrolysis: \( HONH_{3}^{+} + H_{2}O \rightleftharpoons HONH_{2} + H_{3}O^{+} \) Set up either Kb or Ka since HONH3 is the conjugate acid of HONH2 (our weak base in part a). Using the relationship: \( K_{a} \times K_{b} = K_{w} \) where \( K_{w} = 1 \times 10^{-14} \) We have Kb, so let's find Ka: \( K_{a} = \dfrac{K_w}{K_b} = \dfrac{1 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.1 \times 10^{-7} \) Now, set up the ICE table: | Acid | -H3O | +HONH2 | |----------|-----|--------| | Initial |0.100 M| 0 |0 | | Change |-x | +x | +x | | Equilibrium | 0.100-x| x | x| \[ K_{a} = \dfrac{[H_{3}O^{+}][HONH_{2}]}{[HONH_{3}^+]} = \dfrac{x^2}{0.100 - x} = 9.09 \times 10^{-7} \] Solving for x: \( x^2 = (0.100) \times (9.1 \times 10^{-7}) \) Now, we'll find the pH using: \( pH = -\log_{10} [H_3O^+] = -\log_{10}(x) \)

c. pH of pure H2O

This is pure water, so the concentrations of H3O+ and OH- are equal: \[ [H_{3}O^{+}] = [OH^{-}] \] Since the product of their concentrations equals the ion product constant for water, Kw: \[ K_{w} = [H_{3}O^{+}][OH^{-}] = 1 \times 10^{-14} \] Because [H3O+] = [OH-]: \[ [H_{3}O^{+}]^2 = 1 \times 10^{-14} \] Taking the square root: \[ [H_{3}O^{+}] = 1 \times 10^{-7} \] And finally, find the pH: \[ pH = -\log_{10}[H_{3}O^{+}] = 7 \]

d. pH of the mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl

This is a buffer solution, as it contains a weak base and its conjugate acid. The best way to approach this problem is by using the Henderson-Hasselbalch equation: \[ pH = pK_{a} + \log \dfrac{[Base]}{[Acid]} \] We already have the value of Ka from part b. So, let's calculate pKa: \[ pK_{a} = -\log_{10}K_{a} \] Now, we plug in the concentrations of the base (HONH2) and the acid (HONH3+), and calculate the pH: \[ pH = pK_{a} + \log \dfrac{[HONH_{2}]}{[HONH_{3}^+]} \]

Key concepts

Understanding Weak Bases

Weak bases are substances that do not fully ionize in solution. Instead, they partially react with water, forming their conjugate acids and hydroxide ions. This partial ionization makes them weaker in producing hydroxide ions compared to strong bases.

For example, in the case of HONH₂, a typical reaction looks like this:

• HONH₂ + H₂O ⇌ HONH₃⁺ + OH⁻

The equilibrium constant for this reaction, called the base dissociation constant ( K_b ), reflects how far the reaction proceeds towards products. A small K_b , like 1.1 imes 10^{-8} , indicates very little ionization.
Understanding weak bases helps us predict how they behave in solutions and their effect on pH .

Exploring Buffer Solutions

Buffer solutions resist changes in pH when small amounts of acid or base are added. They are composed of a weak acid or base and its conjugate counterpart.

In part d of the exercise, the solution contains a weak base ( HONH₂ ) and its conjugate acid ( HONH₃Cl ), creating a buffer. The buffer's ability to maintain a stable pH is crucial for biological and chemical systems.

• A buffer works by neutralizing added acids or bases.
• The conjugate pair balances out the pH changes.

Buffer solutions are essential for many applications where maintaining a specific pH is necessary, such as in physiological processes.

The Role of the Water Ionization Constant

In pure water, molecules dissociate slightly to form equal amounts of hydronium ( H₃O⁺ ) and hydroxide ( OH⁻ ) ions. The equilibrium constant for this process, known as the water ionization constant ( K_w ), is 1 imes 10^{-14} at 25°C.

This constant is pivotal for calculating pH in solutions:

• In pure water, [H₃O⁺] = [OH⁻] = 1 imes 10^{-7}
• Thus, pH = 7 , indicating neutrality.

Understanding K_w helps in determining the pH levels of solutions not only in pure water but also in various aqueous systems, forming a foundational concept for acid-base chemistry.

Henderson-Hasselbalch Equation and Its Applications

The Henderson-Hasselbalch equation is a critical tool for estimating the pH of buffer solutions. It relates the pH to the pKa (the negative logarithm of the acid dissociation constant, K_a) and the concentration ratio of the conjugate base to the conjugate acid.

For a solution containing a weak base and its conjugate acid, the equation is:

• \[ pH = pK_a + \log \frac{[Base]}{[Acid]} \]

This equation simplifies pH calculations without extensive ICE tables (Initial, Change, Equilibrium).

Using the Henderson-Hasselbalch equation helps in designing buffer systems for various practical applications, such as in laboratories and biological systems, by accurately controlling the pH levels.