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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 \(1\)?
- A \(55\,mL\,\frac{M}{{10}}\,HCl + 45\,mL\,\frac{M}{{10}}\,NaOH\)
- B \(75\,mL\,\frac{M}{5}\,HCl + 25\,mL\,\frac{M}{5}\,NaOH\)
- C \(100\,mL\,\frac{M}{{10}}\,HCl + 100\,mL\,\frac{M}{{10}}\,NaOH\)
- D \(60\,mL\,\frac{M}{{10}}HCl + 40\,mL\,\frac{M}{{10}}NaOH\)
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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