`100mL CuSO_(4)(aq)` was electrolyzed using inert electrodes by passing `0.965A `till the `pH` of the resulting solution was 1. The solution after electrollysis was neutralized, treated with excess `KI` and titrated with `0.04M Na_(2)S_(2)O_(3)`. Volume of `Na_(2)S_(2)O_(3)` required was `35mL`. Assuming no volume change during electrolysis, calculate:
(a) duration of electrolysis if current efficiency is `80%`
(b) initial concentration `(M)` of `CuSO_(4)`.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Duration of Electrolysis

1. **Determine the concentration of H⁺ ions from pH:**
\[ \text{pH} = 1 \implies [H^+] = 10^{-1} \, \text{M} = 0.1 \, \text{M} \]

2. **Calculate moles of H⁺ in 100 mL of solution:**
\[ \text{Moles of } H^+ = [H^+] \times \text{Volume (L)} = 0.1 \times \frac{100}{1000} = 0.01 \, \text{moles} \]

3. **Determine moles of electrons required:**
The reaction at the anode involves the electrolysis of water, producing H⁺ ions:
\[ 4 \, \text{e}^- \rightarrow 4 \, H^+ + O_2 \]
Therefore, 1 mole of H⁺ requires 1 mole of electrons:
\[ \text{Moles of electrons} = 0.01 \, \text{moles} \]

4. **Calculate total charge (Q) using Faraday's constant (F = 96500 C/mol):**
\[ Q = \text{moles of electrons} \times F = 0.01 \times 96500 = 965 \, \text{C} \]

5. **Use the formula relating charge, current, and time:**
\[ Q = I \times t \implies t = \frac{Q}{I} \]
Given the current efficiency is 80%, the effective current (I_eff):
\[ I_{\text{eff}} = 0.965 \times \frac{80}{100} = 0.772 \, \text{A} \]
Now, substituting the values:
\[ t = \frac{965}{0.772} \approx 1250 \, \text{seconds} \]

### Part (b): Initial Concentration of CuSO₄

1. **Determine the moles of I₂ produced from titration:**
The reaction with Na₂S₂O₃ is:
\[ I_2 + 2 \, Na_2S_2O_3 \rightarrow 2 \, NaI + Na_2S_4O_6 \]
From the titration data:
\[ \text{Millimoles of } Na_2S_2O_3 = \text{Molarity} \times \text{Volume} = 0.04 \, \text{mol/L} \times 35 \, \text{mL} = 1.4 \, \text{mmol} \]
Therefore, moles of I₂:
\[ \text{Moles of } I_2 = \frac{1.4}{2} = 0.7 \, \text{mmol} = 0.0007 \, \text{moles} \]

2. **Relate moles of Cu²⁺ to moles of I₂:**
From the reaction with KI:
\[ CuSO_4 + 2KI \rightarrow 2CuI + I_2 + K_2SO_4 \]
This indicates 1 mole of Cu²⁺ produces 1 mole of I₂. Therefore:
\[ \text{Moles of } Cu^{2+} = 0.0007 \, \text{moles} \]

3. **Calculate the total moles of Cu²⁺:**
The moles of Cu²⁺ produced during electrolysis will be equal to the moles of H⁺ consumed:
\[ \text{Total moles of } Cu^{2+} = 0.0007 + 0.01 = 0.0107 \, \text{moles} \]

4. **Calculate the concentration of CuSO₄:**
\[ \text{Concentration} = \frac{\text{Total moles}}{\text{Volume (L)}} = \frac{0.0107}{0.1} = 0.107 \, \text{M} \]

### Final Answers:
(a) Duration of electrolysis = **1250 seconds**
(b) Initial concentration of CuSO₄ = **0.107 M**