Problem 84 - Chemistry Matter and Change

Question

Metal Cleaner Chromic acid is used as an industrial cleaner for metals. What is \(K_{\mathrm{a}}\) for the second ionization of chromic acid \(\left(\mathrm{H}_{2} \mathrm{CrO}_{4}\right)\) if a 0.040 \(\mathrm{M}\) solution of sodium hydrogen chromate has a pH of 3.946?

Step-by-Step Solution

Answer

The \(K_a\) for the second ionization of chromic acid \(\left(\mathrm{H}_{2} \mathrm{CrO}_{4}\right)\) is approximately \(1.1 \times 10^{-2}\).

Step 1: Write the Ionization Reaction

First, let's write the ionization reaction for the second ionization of chromic acid \((\mathrm{H}_{2} \mathrm{CrO}_{4})\): \[\mathrm{HCrO}_{4}^{-} \rightleftharpoons\mathrm{H}^{+}+ \mathrm{CrO}_{4}^{2-}\]

Step 2: Use Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation to relate the pH, p\(K_a\), and the ratio of the concentrations of the conjugate acid/base pair: \[pH = pK_a + \log{\frac{[\mathrm{CrO}_4^{2-}]}{[\mathrm{HCrO}_4^-]}}\] The pH of the sodium hydrogen chromate solution is given as 3.946, so we can plug that value into the equation: \[3.946 = pK_a + \log{\frac{[\mathrm{CrO}_4^{2-}]}{[\mathrm{HCrO}_4^-]}}\]

Step 3: Convert the Given Concentration

We are given the concentration of sodium hydrogen chromate (\(0.040 M\)). Since sodium hydrogen chromate dissociates completely in water, we can assume that the initial concentration of \(\mathrm{HCrO}_{4}^{-}\) is also \(0.040 M\).

Step 4: Calculate Concentration of H+ ions

Before we move forward, we need to find the concentration of H+ ions in the solution, which can be done using the formula: \[[\mathrm{H}^{+}] = 10^{-pH}\] Using the given pH value of 3.946: \[[\mathrm{H}^{+}] = 10^{-3.946} \approx 1.1 \times 10^{-4} M\]

Step 5: Calculate Equilibrium Concentrations of the Species

Now, we can find the equilibrium concentrations of the species involved in the reaction. At equilibrium: \[[\mathrm{HCrO}_{4}^{-}] = [0.040 M]-[\mathrm{H}^{+}]\] \[[\mathrm{CrO}_{4}^{2-}] = [\mathrm{H}^{+}]\] So: \[[\mathrm{HCrO}_{4}^{-}] = [0.040 M]- 1.1\times10^{-4} M \approx 0.040 M\] \[[\mathrm{CrO}_{4}^{2-}] = 1.1\times10^{-4} M\]

Step 6: Solve for pKa and Ka

Plug the equilibrium concentrations into the Henderson-Hasselbalch equation and solve for p\(K_a\): \[3.946 = pK_a + \log{\left(\frac{1.1\times10^{-4}}{0.040}\right)}\] Solving for p\(K_a\): \[pK_a \approx 3.946 + \log(0.00275) \approx 1.96\] Now, to find \(K_a\) by taking the inverse log of p\(K_a\): \[K_a = 10^{-pK_a} = 10^{-1.96} \approx 1.1 \times 10^{-2}\] So, the \(K_a\) for the second ionization of chromic acid (\(\mathrm{H}_{2}\mathrm{CrO}_{4}\)) is approximately \(1.1 \times 10^{-2}\).

Key Concepts

pH calculation

One of the fundamental concepts in chemistry is the pH calculation, which is critical for understanding the acidity or basicity of a solution. The pH is a scale used to specify how acidic or basic (alkaline) a water-based solution is. It is defined as the negative logarithm (base 10) of the activity of hydronium ions (H₃O⁺) or, more typically, hydrogen ions (H⁺) in a solution. Mathematically, it is represented as:

\[ pH = -\log[H^{+}] \]
In our exercise, the pH of the sodium hydrogen chromate is given as 3.946. We use this value to calculate the concentration of hydrogen ions in the solution by taking the inverse log, leading to the equation:

\[ [H^{+}] = 10^{-pH} \]
This conversion is crucial as it allows us to quantify the acidity and proceed with further calculations in equilibrium chemistry. By establishing the concentration of hydrogen ions, we are setting the stage to explore acid dissociation constants and buffer system behaviors.