Question

A current of $4.0\text{ A}$ is passed through $0.5\text{ L}$ of $0.2\text{ M NaCl}$ solution for $1200\text{s}$. Calculate the $\text{pH}$ of the solution after electrolysis.

• A. \( 1.3 \)
• B. \( 13 \)
• C. \( 7.0 \)
• D. \( 2.0 \)

Hint:

Whenever you notice that the electrolysis of a neutral salt solution releases \(\text{OH}^-\) or \(\text{H}^+\) ions, look immediately at the order of magnitude of the concentration to eliminate choices! Since \([\text{OH}^-] = 0.1\text{ M}\), the solution becomes distinctly basic, which means the \(\text{pH}\) must end up well above 7. This instantly rules out options A, C, and D in under two seconds, making Option B the only physically viable choice.

Answer 1

The Correct Option is B

Step 1 : Understanding the Question:
This problem asks us to determine the pH of a sodium chloride (brine) solution after a specific current has been passed through it for a given duration. During the electrolysis of aqueous sodium chloride, water is reduced at the cathode, producing hydrogen gas and hydroxide ions, while chloride ions are oxidized at the anode to produce chlorine gas. The generation of hydroxide ions increases the basicity of the solution, and we need to calculate the resulting pH.
Step 2 : Key Formulas and Approach:
First, we find the total electrical charge $Q$ passed through the electrolyte using:
\[ Q = I \times t \]
Where $I$ is the current in amperes and $t$ is the time in seconds.
Next, we calculate the number of moles of electrons transferred using Faraday's constant ($F \approx 96500\text{ C/mol}$):
\[ \text{Moles of } e^- = \frac{Q}{F} \]
The reduction of water at the cathode is represented by the half-reaction:
\[ 2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq) \]
This shows that 1 mole of electrons produces 1 mole of hydroxide ions.
We will check whether the chloride ions are the limiting reactant. If not, the amount of $\text{OH}^-$ formed is determined by the total charge. Finally, we calculate the concentration of $\text{OH}^-$, find the pOH, and then find the pH using:
\[ \text{pH} = 14 - \text{pOH} \]
Step 3 : Detailed Explanation:

We start by calculating the total electric charge passed through the solution: $Q = 4.0\text{ A} \times 1200\text{ s} = 4800\text{ Coulombs}$.

Next, we determine the number of moles of electrons that this charge represents: $\text{Moles of } e^- = \frac{4800\text{ C}}{96500\text{ C/mol}} \approx 0.04974\text{ moles} \approx 0.05\text{ moles}$.

Now, we check the initial quantity of sodium chloride in the solution to ensure it is not a limiting reactant: $\text{Initial moles of NaCl} = \text{Molarity} \times \text{Volume} = 0.2\text{ mol/L} \times 0.5\text{ L} = 0.1\text{ moles}$.

Since the reaction requires $0.05\text{ moles}$ of electrons to oxidize $0.05\text{ moles}$ of chloride ions, and we have $0.1\text{ moles}$ available, the chloride ions are in excess, and the reaction is limited only by the electricity passed.

From the stoichiometry of the cathode reaction, the moles of hydroxide ions produced are equal to the moles of electrons transferred. Therefore, $\text{Moles of OH}^- \text{ produced} = 0.05\text{ moles}$.

We calculate the molar concentration of the hydroxide ions in the $0.5\text{ L}$ solution: $[\text{OH}^-] = \frac{\text{Moles of OH}^-}{\text{Volume of Solution}} = \frac{0.05\text{ moles}}{0.5\text{ L}} = 0.1\text{ M} = 10^{-1}\text{ M}$.

Using the definition of pOH, we get: $\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(10^{-1}) = 1$.

Finally, we calculate the pH of the solution at standard temperature ($25^\circ\text{C}$): $\text{pH} = 14 - \text{pOH} = 14 - 1 = 13$.

Step 4 : Final Answer:
The pH of the solution after electrolysis is 13, which corresponds to Option (B).