Question

Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?

• A. 100 mL \(\frac{M}{10}\) HCl + 100 mL \(\frac{M}{10}\) NaOH
• B. 75 mL \(\frac{M}{5}\) HCl + 25 mL \(\frac{M}{5}\) NaOH
• C. 60 mL \(\frac{M}{10}\) HCl + 40 mL \(\frac{M}{10}\) NaOH
• D. 55 mL \(\frac{M}{10}\) HCl + 45 mL \(\frac{M}{10}\) NaOH

Hint:

Subtract the moles of NaOH from the moles of HCl; if HCl is in excess, [H⁺] = excess moles / total volume. pH = 1 needs [H⁺] = 0.1 M.

Answer 1

The Correct Option is B

For each mixture compute the excess H⁺ (in mmol) and divide by the total volume.
(1) 100 × 0.1 = 10 mmol HCl, 10 mmol NaOH: neutral, pH = 7.
(2) HCl: 75 × 0.2 = 15 mmol; NaOH: 25 × 0.2 = 5 mmol. Excess HCl = 10 mmol in 100 mL: [H⁺] = 0.1 M, pH = 1.
(3) HCl: 6 mmol; NaOH: 4 mmol. Excess = 2 mmol in 100 mL: [H⁺] = 0.02 M, pH = 1.7.
(4) HCl: 5.5 mmol; NaOH: 4.5 mmol. Excess = 1 mmol in 100 mL: [H⁺] = 0.01 M, pH = 2.
Only (2) gives pH = 1.
Answer: (2) 75 mL \(\frac{M}{5}\) HCl + 25 mL \(\frac{M}{5}\) NaOH