Ksp of Al(OH)3 is 1.0×10−15. pH of the saturated solution is about

Al(OH)3⇌Al3++3OH− S 3S∴ Ksp=Al3+OH−3⇒1×10−15=27S4 S=1×10−152714=7.8×10−5 ∴ OH−=3S=2.34×10−4 ∴ pOH=−logOH−=3.63 ∴ pH=10.37≈10.4

Ksp of Al(OH)3 is 1.0×10−15. pH of the saturated solution is about

Al(OH)3⇌Al3++3OH− S 3S∴ Ksp=Al3+OH−3⇒1×10−15=27S4 S=1×10−152714=7.8×10−5 ∴ OH−=3S=2.34×10−4 ∴ pOH=−logOH−=3.63 ∴ pH=10.37≈10.4

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$K_{sp} of Al (OH)_{3} is 1 .0 \times 10^{- 15} .$ pH of the saturated solution is about

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9.0

5.0

10.4

4.1

answer is D.

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Detailed Solution

$Al (OH)_{3} ⇌ {Al}^{3 +} + 3 {OH}^{-} S 3 S$

$∴ K_{sp} = [{Al}^{3 +}] {[{OH}^{-}]}^{3} \Rightarrow 1 \times 10^{- 15} = 27 S^{4} S = {(\frac{1 \times 10^{- 15}}{27})}^{\frac{1}{4}} = 7 .8 \times 10^{- 5} ∴ [{OH}^{-}] = 3 S = 2 .34 \times 10^{- 4} ∴ pOH = - \log [{OH}^{-}] = 3 .63 ∴ pH = 10 .37 \approx 10 .4$