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 strong acid-strong base titrations, always focus on the moles of $H^+$ and $OH^-$. The excess ion determines the final pH. Remember to calculate concentrations using the total volume after mixing.

Answer 1

The Correct Option is A

Step 1: Take the mixing ratio literally as actual volumes: 2 L of the 0.02 M NaOH solution and 3 L of the 0.01 M HCl solution. Moles of \(NaOH = 0.02 \times 2 = 0.04\) mol, moles of \(HCl = 0.01 \times 3 = 0.03\) mol.

Step 2: Since \(NaOH\) and \(HCl\) neutralise in a 1:1 ratio, the \(0.03\) mol of \(HCl\) is completely used up, leaving \(0.04 - 0.03 = 0.01\) mol of \(OH^-\) unreacted in a total volume of \(2 + 3 = 5\) L.

Step 3: Concentration of leftover \(OH^-\) is \(\frac{0.01}{5} = 0.002\) M, so \(pOH = -\log(0.002) \approx 2.7\).

Step 4: Using \(pH + pOH = 14\), \(pH = 14 - 2.7 = 11.3\).
\[ \boxed{pH \approx 11.3} \]