Question

100 mL of 0.04 N HCl aqueous solution is mixed with 100 mL of 0.02 N NaOH solution. The pH of the resulting solution is:

• A. 1.0
• B. 1.7
• C. 2.0
• D. 2.3

Hint:

When mixing an acid and a base, the pH of the resulting solution depends on the remaining concentration of H\(^+\) or OH\(^-\) after neutralization. The pH can be calculated using the formula: \[ \text{pH} = \log [\text{H}^+]. \]

Answer 1

The Correct Option is C

Step 1: Understanding the Problem 
We are mixing 100 mL of 0.04 N HCl with 100 mL of 0.02 N NaOH. We need to find the pH of the resulting solution. 
Step 2: Calculating the Moles of H\(^+\) and OH\(^-\) 
Moles of H\(^+\) from HCl: \[ \text{Moles of H}^+ = 0.04 \, \text{N} \times 0.1 \, \text{L} = 0.004 \, \text{moles}. \] 
Moles of OH\(^-\) from NaOH: \[ \text{Moles of OH}^-= 0.02 \, \text{N} \times 0.1 \, \text{L} = 0.002 \, \text{moles}. \] 
Step 3: Determining the Remaining Moles of H\(^+\) 
The reaction between H\(^+\) and OH\(^-\) is: \[ \text{H}^+ + \text{OH}^-\rightarrow \text{H}_2\text{O}. \] 
Moles of H\(^+\) remaining after the reaction: \[ \text{Remaining moles of H}^+ = 0.004 - 0.002 = 0.002 \, \text{moles}. \] 
Step 4: Calculating the Concentration of H\(^+\) 
Total volume of the resulting solution: \[ 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L}. \] 
Concentration of H\(^+\): \[ [\text{H}^+] = \frac{0.002 \, \text{moles}}{0.2 \, \text{L}} = 0.01 \, \text{M}. \] 
Step 5: Calculating the pH 
The pH is given by: \[ \text{pH} = \log [\text{H}^+]. \] 
Substituting the concentration of H\(^+\): \[ \text{pH} = \log (0.01) = 2.0. \] 
Step 6: Matching with the Options 
The calculated pH is 2.0, which corresponds to option (C). Final Answer: The pH of the resulting solution is (C) 2.0.