COMEDK · Chemistry · 18. Electrochemistry
Assuming a theoretical setup where only the anodic removal of \(\text{OH}^-\) affects the solution, and no change in volume, the time required to obtain a solution of \(\text{pH} = 4\) by electrolysis of \(100\text{ mL}\) of \(0.1\text{ M}\) \(\text{NaOH}\) (using current \(0.5\text{ A}\)) will be:
- A \(1.93 \times 10^3\text{ s}\)
- B \(2.63 \times 10^3\text{ s}\)
- C \(4.26 \times 10^3\text{ s}\)
- D \(1.80 \times 10^3\text{ s}\)
Answer & Solution
Correct Answer
(A) \(1.93 \times 10^3\text{ s}\)
Step-by-step Solution
Detailed explanation
Initial moles of \(OH^- = M \times V = 0.1 \times 0.1 = 0.01\) mol Target pH \(= 4\) \(\Rightarrow\) \([H^+] = 10^{-4}\) M, meaning solution is acidic \(\Rightarrow\) practically all \(OH^-\) must be removed. Moles of \(OH^-\) to be removed \(\approx 0.01\) mol Anodic reaction:…
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