Question

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

Short answer

The pH of the \(0.050 \mathrm{M}\) \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}\) solution is approximately 11.33.

Step-by-step solution

Write down the ionization equation

Ethylene diamine acts as a base in the solution, where its ionization equation can be represented as follows: \[\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}_{2}^{+} + \mathrm{OH}^{-}\]

Set up the ICE table

It's essential to use an ICE (initial, change, equilibrium) table to organize the changing concentrations during the ionization process: \[ \begin{array}{c|c|c|c} & (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH} & (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+} & \mathrm{OH}^{-} \\ \hline I & 0.050 & 0 & 0 \\ C & -x & +x & +x \\ E & 0.050-x & x & x \\ \end{array} \] Where \(I\) indicates initial concentrations, \(C\) is the change in concentrations, and \(E\) represents equilibrium concentrations.

Write the expression for \(K_{b}\) and substitute the equilibrium concentrations

The base dissociation constant (\(K_{b}\)) expression can be written as follows: \[K_{b} = \frac{[\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}_{2}^{+}][\mathrm{OH}^{-}]}{[\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}]}\] Plug in the equilibrium concentrations from the ICE table: \[K_{b} = \frac{x \cdot x}{0.050 - x} = \frac{x^{2}}{0.050 - x}\]

Solve for \(x\) using the given \(K_{b}\) value

Now, we need to insert the given value of \(K_{b}\) and solve for \(x\): \[1.3 \times 10^{-3} = \frac{x^{2}}{0.050 - x}\] This equation can be simplified further: \[x^{2} + 1.3 \times 10^{-3}x - 0.050 \times 1.3 \times 10^{-3} = 0\] The quadratic equation is difficult to solve by hand, so a calculator can be used to find the value of \(x\): \[x \approx 0.00214\] This value, \(x\), represents the concentration of \(\mathrm{OH}^{-}\) at equilibrium.

Calculate the \(\mathrm{pOH}\) and then the \(\mathrm{pH}\)

Now, we can calculate the pOH using the \(\mathrm{OH}^{-}\) concentration: \[\mathrm{pOH} = -\log_{10}[\mathrm{OH}^{-}] = -\log_{10}(0.00214) \approx 2.67\] Finally, we can find the pH of the solution by subtracting the pOH from 14: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.67 \approx 11.33\] So, the \(\mathrm{pH}\) of the \(0.050 \mathrm{M}\) \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2}\mathrm{NH}\) solution is approximately 11.33.

Key concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is key to calculating pH in solutions. This concept involves the balance between acids, bases, and the ions they release into solutions. For our example, ethylene diamine (\((\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}\)) is a base. It reacts with water to produce hydroxide ions (OH\(^-\)) and a corresponding conjugate acid. This process helps us understand how the concentration of different ions changes and reaches equilibrium over time.In such equilibrium, the forward and reverse reactions occur at the same rate. This means the concentrations of reactants and products remain constant. Knowing this helps us determine the solution's pH, as ion concentrations are vital in these calculations.

Base Dissociation Constant

The base dissociation constant, denoted as \(K_b\), measures the strength of a base in solution. It tells us how well a base can dissociate to form ions. In our exercise, we used the equation:\[K_b = \frac{[(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}_2^+][\mathrm{OH}^-]}{[(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}]}\]Here, \(K_b\) represents the equilibrium state of the base gene \((\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}\). A higher \(K_b\) value denotes a stronger base that dissociates more in water, producing more hydroxide ions.to calculating pH, we utilize the value of \(K_b\) to derive the concentration of OH\(^-\). This helps us connect with the pOH and then the pH of the solution.

ICE Table

An ICE table (Initial, Change, Equilibrium) is a simple tool to track concentration changes in a chemical reaction.Let's look at our exercise:- **Initial**: Start concentrations before the reaction (e.g., 0.050 M of \((\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}\), 0 for others)- **Change**: Represents what's gained (+x) or lost (-x) during the reaction.- **Equilibrium**: Final concentrations after the reaction reaches equilibrium.The ICE table helps us organize information clearly. It's especially useful when inserting values into the dissociation constant expression to find unknown concentrations. For step-by-step equilibrium calculations, this method is invaluable.

Hydroxide Ion Concentration

The concentration of hydroxide ions \((\mathrm{OH}^-)\) plays a crucial role in determining a solution's pH. In our calculation, using the ICE table, we determine \(x\) represents \(\mathrm{OH}^-\) at equilibrium.To express this scientifically,- \[\text{Concentration of } \mathrm{OH}^- = x = 0.00214\, \text{M}\]Once we know \([\mathrm{OH}^-]\), finding the pOH becomes straightforward:- \[\mathrm{pOH} = -\log_{10}([\mathrm{OH}^-])\]Finally, translating from pOH to pH requires simple subtraction:- \[\mathrm{pH} = 14 - \mathrm{pOH}\]This sequence is essential. It connects the chemical groundwork to the desired pH result of 11.33.