25.0 mL of 0.050 M Ba(NO3)2 is mixed with 25.0 mL of 0.020 M NaF.Ksp of BaF2 is 0.5 x 10–6 at 298 K. — JEE Mains Chemistry Past Papers Chemistry Question
Question
25.0 mL of 0.050 M Ba(NO3)2 is mixed with 25.0 mL of 0.020 M NaF.Ksp of BaF2 is 0.5 x 10–6 at 298 K. The ratio of [Ba2+] [F–]2 and Ksp is (Nearest integer)
Answer: .
💡 Solution & Explanation
On mixing equal volume concentration is Half. [Ba2+] = 05 . = 0.025 [F–] = 02 . = 0.01 KIP[BaF2] = [Ba2+] [F–]2 = 2.5 x 10–2 x (10–2)2 = 2.5 x 10–6 as KIP KSP so precipitation occur. as solution is super saturated so maximum possible value of KIP is equal to KSP so SP K ] F [] Ba [ = 1
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