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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

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 \(1\)?

  1. A \(55\,mL\,\frac{M}{{10}}\,HCl + 45\,mL\,\frac{M}{{10}}\,NaOH\)
  2. B \(75\,mL\,\frac{M}{5}\,HCl + 25\,mL\,\frac{M}{5}\,NaOH\)
  3. C \(100\,mL\,\frac{M}{{10}}\,HCl + 100\,mL\,\frac{M}{{10}}\,NaOH\)
  4. D \(60\,mL\,\frac{M}{{10}}HCl + 40\,mL\,\frac{M}{{10}}NaOH\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(75\,mL\,\frac{M}{5}\,HCl + 25\,mL\,\frac{M}{5}\,NaOH\)

Step-by-step Solution

Detailed explanation

\(75\,mL\,\frac{M}{5}\,HCl + 25\,ml\,\frac{M}{5}\,NaOH\) \(25\,mL\,\frac{M}{5}\,NaOH\,\) will neutralise \(25\,mL\,\frac{M}{5}\,HCl\) \(75 - 25 = 50\,mL\,\frac{M}{5}\,HCl\) will remain. Total volume will be \(75 + 25 = 100\,mL\) \(50\,mL\,\frac{M}{5}\,HCl\) is diluted to…
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