JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(25.0\,mL\) of \(0.050\,M\,Ba \left( NO _3\right)_2\) is mixed with \(25.0\,mL\) of \(0.020\,M\) \(NaF\),\(K _{ sp }\) of \(BaF _2\) is \(0.5 \times 10^{-6}\) at \(298\,K\). The ratio of \(\left[ Ba ^{2+}\right]\left[F^{-}\right]^2\) and \(K _{ sp }\) is \(.......\). (Nearest integer)
- A \(2\)
- B \(3\)
- C \(5\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\({\left[ Ba ^{+2}\right]=\frac{25 \times 0.05}{50}=0.025 M }\) \({\left[ F ^{-}\right]=\frac{25 \times 0.02}{50}=0.01 M }\) \({\left[ Ba ^{+2}\right]\left[ F ^{-}\right]^2=25 \times 10^{-7}}\) \(K _{ sp }=5 \times 10^{-7} \text { (given) }\)…
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