Question

What is the pH of the resulting solution when equal volumes of $0.1\, M \,NaOH$ and $0.01\, M$ $HCl$ are mixed ?

• A. 7
• B. 1.04
• C. 12.65
• D. 2

Answer 1

The Correct Option is C

To determine the pH of the resulting solution when equal volumes of 0.1 \, M \, NaOH and 0.01 \, M \, HCl are mixed, follow these steps:

1. First, calculate the number of moles of NaOH and HCl. Assume the volume of each solution is V liters for simplicity.
2. The number of moles of NaOH is given by:
   \text{Moles of } NaOH = 0.1 \, \text{M} \times V = 0.1V
   The number of moles of HCl is given by:
   \text{Moles of } HCl = 0.01 \, \text{M} \times V = 0.01V
3. Since the solutions are mixed in equal volumes, the reaction between NaOH and HCl will proceed as follows:
   NaOH + HCl \rightarrow NaCl + H_2O
   This is a neutralization reaction.
4. Calculate the remaining moles of NaOH after the reaction: - Moles of NaOH left = Initial moles of NaOH - Moles of HCl reacting - Moles of NaOH left = 0.1V - 0.01V = 0.09V Thus, 0.09V moles of NaOH remain unreacted.
5. Since NaOH is a strong base, we will calculate the concentration of the OH^− ions: - Total volume of the mixed solution = V + V = 2V - Concentration of OH^− ions = \frac{0.09V}{2V} = 0.045 \, M
6. Calculate the pOH of the solution:
   \text{pOH} = -\log[OH^-] = -\log(0.045) \approx 1.35
7. Finally, calculate the pH:
   \text{pH} = 14 - \text{pOH} = 14 - 1.35 = 12.65

Hence, the pH of the resulting solution is approximately 12.65.