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11/10/2021 10:53 am
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The solubility of PbI2 at 25°C is 0.7 g L-1. The solubility product of PbI2 at this temperature is (molar mass of PbI2 = 461.2 g mol-1)
(a) 1.40 × 10-9
(b) 0.14 × 10-9
(c) 140 × 10-9
(d) 14.0 × 10-9
1 Answer
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11/10/2021 11:03 am
Answer: (d) 14.0 × 10-9
PbI2 ⇌ Pb++ + 2I-
s 2s
Ksp = s x (2s)2 = 4s3
= 4 x \(\Big(\frac{0.7}{46.12}\Big)^2\)
= 14.0 x 10-9
This post was modified 3 years ago 2 times by Mayank1444
This post was modified 3 years ago 2 times by admin