A current of 0.20A is passed for 482 s through 50.0mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis?

Options

A. 0.0159M
B. 0.0199M
C. 0.10M
D. 0.030M

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the total charge (Q)
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)

Given:
- \( I = 0.20 \, A \)
- \( t = 482 \, s \)

Calculating:
\[ Q = 0.20 \, A \times 482 \, s = 96.4 \, C \]

### Step 2: Calculate the number of Faradays (n)
To find the number of Faradays, we use the relationship:
\[ n = \frac{Q}{F} \]
where:
- \( F \) = Faraday's constant (\( 96500 \, C/mol \))

Calculating:
\[ n = \frac{96.4 \, C}{96500 \, C/mol} = 9.98 \times 10^{-4} \, mol \]

### Step 3: Determine the moles of hydroxide ions (OH⁻)
From the electrolysis reaction of NaCl, we know that:
\[ 2 \, \text{NaCl} \rightarrow 2 \, \text{Na}^+ + 2 \, \text{OH}^- + \text{H}_2 + \text{Cl}_2 \]

This indicates that 2 moles of NaCl produce 2 moles of OH⁻. Therefore, the moles of OH⁻ produced will be equal to the moles of NaCl reacted.

Since 2 moles of electrons produce 2 moles of OH⁻, the moles of OH⁻ produced will be:
\[ \text{Moles of OH}^- = n = 9.98 \times 10^{-4} \, mol \]

### Step 4: Calculate the concentration of hydroxide ions (OH⁻)
The concentration (C) of hydroxide ions can be calculated using the formula:
\[ C = \frac{\text{Number of moles of OH}^-}{\text{Volume in liters}} \]

Given:
- Volume = 50.0 mL = 0.050 L

Calculating:
\[ C = \frac{9.98 \times 10^{-4} \, mol}{0.050 \, L} = 0.01996 \, mol/L \]

### Final Answer
The hydroxide ion concentration in the solution after electrolysis is approximately:
\[ \text{[OH}^-] \approx 0.0200 \, M \]

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