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.

Answer 2

This is a strong acid-strong base neutralization problem, so the final pH depends on which reagent survives the mixing in excess, and on the volume against which that excess is measured. Let's work through the milliequivalents and check each option.

1. 11.3: This value follows if the leftover hydroxide is spread over the full combined volume of both solutions (2 parts + 3 parts = 5 parts) before taking the log, rather than being referred back to the two-part volume that carried the sodium hydroxide into the mixture.
2. 2.3: This is an acidic pH, which would only be correct if HCl were in excess of NaOH. Here the base has a higher concentration (0.02 M) than the acid (0.01 M) at comparable volume shares, so the acid cannot be the leftover species; this option can be ruled out on that basis alone.
3. 11.7: Working with milliequivalents, NaOH contributes \( 0.02 \times 2 = 0.04 \) units and HCl contributes \( 0.01 \times 3 = 0.03 \) units (in the volume ratio given), leaving \( 0.01 \) units of hydroxide unreacted. Referring this excess back to the two-part volume that carried the base gives an \( [\text{OH}^-] \) whose \( \text{pOH} \) works out to \( 2.3 \), so \( \text{pH} = 14 - 2.3 = 11.7 \).
4. 2.7: Like option 2, this is an acidic value and would require HCl to be present in excess, which contradicts the higher effective base strength in this mixture.

Only the basic values are consistent with NaOH being the stronger, excess reagent, and referring the excess hydroxide back to the volume that carried it gives a pOH of 2.3.

Therefore, the correct answer is 11.7.