Question

A 0.02(M) NaOH solution and a 0.01(M) HCl solution are mixed in the volume ratio of 2:3. The pH of the mixed solution will be

• A. 11.3
• B. 2.3
• C. 11.7
• D. 2.7

Hint:

For neutralization problems, calculate the excess concentration of ions to find the pH or pOH.

Answer 1

The Correct Option is C

Step 1: Understanding the neutralization.
When NaOH (a strong base) and HCl (a strong acid) are mixed, they neutralize each other. The number of moles of each reactant determines the excess concentration of hydroxide or hydrogen ions in the solution. Step 2: Calculating the excess concentration.
Let the volume of NaOH be \( 2x \) and the volume of HCl be \( 3x \). The moles of NaOH will be \( 0.02 \times 2x \) and for HCl, it will be \( 0.01 \times 3x \). After neutralization, the excess amount of OH- will be: \[ \text{Excess OH-} = 0.02 \times 2x - 0.01 \times 3x = 0.04x - 0.03x = 0.01x \] Thus, the pH of the solution is determined by the excess concentration of OH- ions. Step 3: Conclusion.
The pH of the solution is calculated using the formula for pOH and pH: \[ \text{pOH} = -\log [\text{OH}^-] \] Substituting the values gives a pH of 11.7. Step 4: Conclusion.
The correct answer is (3) 11.7.