Question

A solution is made by mixing 100 mL of 0.1 M HCl and 100 mL of 0.2 M NaOH. pH of resulting solution is:

• A. 12.0
• B. 11.0
• C. 7.0
• D. 12.7

Hint:

When mixing strong acids and bases, always first calculate the moles of each reactant to determine the excess.

Answer 1

The Correct Option is D

pH of Resulting Solution: HCl + NaOH 

Step 1: Moles of Reactants

\[ \text{HCl: } 0.1\ \text{mol/L} \times 0.1\ \text{L} = 0.01\ \text{mol} \] \[ \text{NaOH: } 0.2\ \text{mol/L} \times 0.1\ \text{L} = 0.02\ \text{mol} \]

Step 2: Reaction

\[ \text{HCl + NaOH} \rightarrow \text{NaCl + H}_2\text{O} \] \[ \text{Excess NaOH} = 0.02 - 0.01 = 0.01\ \text{mol} \]

Step 3: Concentration of OH⁻

\[ [\text{OH}^-] = \frac{0.01}{0.2} = 0.05\ \text{M} \]

Step 4: pOH and pH

\[ \text{pOH} = -\log(0.05) \approx 1.3 \] \[ \text{pH} = 14 - 1.3 = 12.7 \]

✅ Final Answer:

\[ \boxed{\text{pH} \approx 12.7} \]