JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The \(OH^-\) concentration in a mixture of \(5.0\, \mathrm{~mL}\) of \(0.0504\, \mathrm{M} \,\mathrm{NH}_{4} \mathrm{Cl}\) and \(2\, \mathrm{~mL}\) of \(0.0210 \,\mathrm{M}\, \mathrm{NH}_{3}\) solution is \(\mathrm{x} \,\times 10^{-6}\, \mathrm{M}\). The value of \(\mathrm{x}\) is ..... . (Nearest integer) \(\left[\right.\) Given \(\mathrm{K}_{m}=1 \times 10^{-14}\) and \(\left.\mathrm{K}_{\mathrm{b}}=1.8 \times 10^{-5}\right]\)
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{NH}_{4}^{+}\right]=0.0504 \,\&\,\left[\mathrm{NH}_{3}\right]=0.0210\) So \(\mathrm{K}_{b}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{HO}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}\)…
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