At 25C o the KSP values of Ni (OH)2 and Cd(OH)2 are 2.0×10−15 and 2.5×10−14 respectively. If S1 and S2 are respectively their molar solubilities, then the ratio of S1 to S2 is

4s13=2×10-15s13=5×10-164s23=2.5×10-14s23=62.5×10-16s1s23=5×10-1662.5×10-16=0.08s1s2=5×10-1662.5×10-1613=0.08s1s2=8100=225

At 25C o the KSP values of Ni (OH)2 and Cd(OH)2 are 2.0×10−15 and 2.5×10−14 respectively. If S1 and S2 are respectively their molar solubilities, then the ratio of S1 to S2 is

4s13=2×10-15s13=5×10-164s23=2.5×10-14s23=62.5×10-16s1s23=5×10-1662.5×10-16=0.08s1s2=5×10-1662.5×10-1613=0.08s1s2=8100=225

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At $25 {}^{o}C$ the K_SPvalues of $Ni {(OH)}_{2}$ and $Cd {(OH)}_{2}$ are $2.0 \times 10^{- 15}$ and $2.5 \times 10^{- 14}$ respectively. If S₁ and S₂ are respectively their molar solubilities, then the ratio of S₁ to S₂ is

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3 : 8

4 : 3

25 : 2

2 : 25

answer is D.

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$4 s_{1}^{3} = 2 \times 10^{- 15}$

$s_{1}^{3} = 5 \times 10^{- 16}$

$4 s_{2}^{3} = 2.5 \times 10^{- 14}$

$s_{2}^{3} = 62.5 \times 10^{- 16}$

${[\frac{s_{1}}{s_{2}}]}^{3} = [\frac{5 \times 10^{- 16}}{62.5 \times 10^{- 16}}] = 0.08$

$[\frac{s_{1}}{s_{2}}] = {[\frac{5 \times 10^{- 16}}{62.5 \times 10^{- 16}}]}^{\frac{1}{3}} = 0.08$

$[\frac{s_{1}}{s_{2}}] = \frac{8}{100} = \frac{2}{25}$

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At 25C o the KSP values of Ni (OH)2 and Cd(OH)2 are 2.0×10−15 and 2.5×10−14 respectively. If S1 and S2 are respectively their molar solubilities, then the ratio of S1 to S2 is

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At $25 {}^{o}C$ the K_SPvalues of $Ni {(OH)}_{2}$ and $Cd {(OH)}_{2}$ are $2.0 \times 10^{- 15}$ and $2.5 \times 10^{- 14}$ respectively. If S₁ and S₂ are respectively their molar solubilities, then the ratio of S₁ to S₂ is