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QNEET-2018-Ch-872: NEET 2018 Chemistry - Class 11 NEET-2018-Ch-872 Solution: The solubility of \(BaSO_4\) in water is \(2.42 10^{-3}\ {g…

The solubility of BaSO4BaSO_4 in water is 2.42×10−3 g L−12.42 \times 10^{-3}\ \text{g L}^{-1} at 298 K. The value of its solubility product (KspK_{sp}) will be (Given molar mass of BaSO4BaSO_4 = 233 g mol−1mol^{-1}) 1.08×10−10 mol2 L−21.08 \times 10^{-10}\ \text{mol}^2\ \text{L}^{-2} 1.08×10−12 mol2 L−21.08 \times 10^{-12}\ \text{mol}^2\ \text{L}^{-2} 1.08×10−14 mol2 L−21.08 \times 10^{-14}\ \text{mol}^2\ \text{L}^{-2} 1.08×10−8 mol2 L−21.08 \times 10^{-8}\ \text{mol}^2\ \text{L}^{-2}

Multiple Choice Options

  • A 1.08 _ 10 �10 mol 2 L �2
  • B 1.08 _ 10 �12 mol 2 L �2
  • C 1.08 _ 10 �14 mol 2 L �2
  • D 1.08 _ 10 �8 mol 2 L �2

Step-by-Step Verified Solution

Given, Solubility of BaSO 4 = 2.42 _ 10 �3 g L �1 Convert solubility in mol/lit. s = 2.42 _ 10 _ 3 233 = 1.04 _ 10 _ 5 mol L -1 .tg {border-collapse:collapse;border-color:#ccc;border-spacing:0;} .tg td{background-color:#fff;border-color:#ccc;border-style:solid;border-width:0px;color:#333; font-family:Arial, sans-serif;font-size:14px;overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{background-color:#f0f0f0;border-color:#ccc;border-style:solid;border-width:0px;color:#333; font-family:Arial, sans-serif;font-size:14px;font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-p8sp{border-color:inherit;font-size:20px;text-align:center;vertical-align:top} BaSO4(s) | _ | Ba 2+ (aq) | + | SO 4 2- (aq) s | s K sp = [Ba 2+ ][SO 4 2- ] = s 2 = ( 1.04 _ 10 _ 5 ) 2 = 1.08 _ 10 -10 mol 2 L �2

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