Question

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :

1. \(60ml\frac{M}{10}HCL+40ml\frac{M}{10}NaOH\)
2. \(55ml\frac{M}{10}HCL+45ml\frac{M}{10}NaOH\)
3. \(75ml\frac{M}{5}HCL+25ml\frac{M}{5}NaOH\)
4. \(100ml\frac{M}{10}HCL+100ml\frac{M}{10}NaOH\)

pH of which one of them will be equal to 1 ?

• A. b
• B. d
• C. a
• D. c

Answer 1

The Correct Option is D

 The problem involves four solutions made by mixing different volumes of hydrochloric acid (HCl) and sodium hydroxide (NaOH). We need to determine which solution has a pH equal to 1.

The pH of a solution is calculated based on the concentration of hydrogen ions \((H^+)\) present in the solution. Since HCl is a strong acid, it completely dissociates in water to produce \(H^+\) ions. The concentration of \(H^+\) ions gives us the pH:

\(pH = -\log[H^+]\)

The solutions are:

1. \(60ml \frac{M}{10} \text{ HCl} + 40ml \frac{M}{10} \text{ NaOH}\)
2. \(55ml \frac{M}{10} \text{ HCl} + 45ml \frac{M}{10} \text{ NaOH}\)
3. \(75ml \frac{M}{5} \text{ HCl} + 25ml \frac{M}{5} \text{ NaOH}\)
4. \(100ml \frac{M}{10} \text{ HCl} + 100ml \frac{M}{10} \text{ NaOH}\)

Let us evaluate each option:

1. For solution (a):
   • Number of moles of HCl = Volume (L) × Molarity = \(0.060 \times 0.1 = 0.006 \text{ moles}\)
   • Number of moles of NaOH = \(0.040 \times 0.1 = 0.004 \text{ moles}\)
   • After reaction, \(0.006 - 0.004 = 0.002 \text{ moles HCl}\) remain.
   • Total volume = \(60ml + 40ml = 100ml = 0.1L\)
   • Concentration of \(H^+\) = \(\frac{0.002}{0.1} = 0.02 \text{ M}\)
   • \(pH = -\log(0.02) \approx 1.7\)
2. For solution (b):
   • Number of moles of HCl = \(0.055 \times 0.1 = 0.0055 \text{ moles}\)
   • Number of moles of NaOH = \(0.045 \times 0.1 = 0.0045 \text{ moles}\)
   • After reaction, \(0.0055 - 0.0045 = 0.001 \text{ moles HCl}\) remain.
   • Total volume = \(55ml + 45ml = 100ml = 0.1L\)
   • Concentration of \(H^+\) = \(\frac{0.001}{0.1} = 0.01 \text{ M}\)
   • \(pH = -\log(0.01) = 2\)
3. For solution (c):
   • Number of moles of HCl = \(0.075 \times 0.2 = 0.015 \text{ moles}\)
   • Number of moles of NaOH = \(0.025 \times 0.2 = 0.005 \text{ moles}\)
   • After reaction, \(0.015 - 0.005 = 0.01 \text{ moles HCl}\) remain.
   • Total volume = \(75ml + 25ml = 100ml = 0.1L\)
   • Concentration of \(H^+\) = \(\frac{0.01}{0.1} = 0.1 \text{ M}\)
   • \(pH = -\log(0.1) = 1\)
4. For solution (d):
   • Number of moles of HCl = \(0.100 \times 0.1 = 0.01 \text{ moles}\)
   • Number of moles of NaOH = \(0.100 \times 0.1 = 0.01 \text{ moles}\)
   • NaOH and HCl completely neutralize each other.
   • No \(H^+\) ions are present, so the solution is neutral with \(pH = 7\).

Thus, the solution with pH equal to 1 is option c.