JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
A litre of buffer solution contains \(0.1\) mole of each of \(NH _3\) and \(NH _4 Cl\). On the addition of \(0.02\,mole\) of \(HCl\) by dissolving gaseous \(HCl\), the \(pH\) of the solution is found to be \(......\times 10^{-3}(\) Nearest integer) \(\left[\right.\) Given : \(pK _{ b }\left( NH _3\right)=4.745\) \(\log 2=0.301\) \(\log 3=0.477\) \(T =298\,K ]\)
- A \(9080\)
- B \(9079\)
- C \(9081\)
- D \(9082\)
Answer & Solution
Correct Answer
(B) \(9079\)
Step-by-step Solution
Detailed explanation
In resultant solution \(n _{ NH _3}=0.1-0.02=0.08\) \(n _{ NH _4 Cl }= n _{ NH _4^{+}}=0.1+0.02=0.12\) \(pOH = pK _{ b }+\log \frac{\left[ NH _4^{+}\right]}{\left[ NH _3\right]}\) \(=4.745+\log \frac{0.12}{0.08}\) \(=4.745+\log \frac{3}{2}\) \(=4.745+0.477-0.301\) \(pOH =4.921\)…
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