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Shweta Thakur posted an Question
November 22, 2020 • 19:16 pm 30 points
  • IIT JAM
  • Chemistry (CY)

In the pure water the answer should be 2 right ???but the answer given in book is 6 .6 * 10^ -6 ple 7. given that the solubility product of radium su

2 n O-10 M NaOH. The i M NaOH. The ionic product of Ni (OH), is 2-0 x 10-15 Un)2 12 m 0-1 NCBRT Solved Example 7. Given that the solubility product of radium sulphate (RaSO) is 4 x 10- Calculate the solubility in (a) pure water (b) 0-10 M Na,SO4 ANSWERS 3.5 x 10-1l mol L- 4. 1-0x 10 mol L-l 1. 1.26x 10 mol L- 5. S (Ni(OH),) = 5-8 x 10- M, S (AgCN) = 7-8 x 10 M, Ni (OH), is more soluble 2. 3.5 x 104

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