What is the pH of a saturated solution at 2 5 ∘ C of slaked lime, Ca ( OH ) 2 , if its K s p is 3.2 × 1 0 − 5 ? [ lo g 2 = 0.3 ]

K s p = 4 s 3 = 3.2 × 1 0 − 5 , so s 3 = 8 × 1 0 − 6 and s = 2 × 1 0 − 2 M . [ OH − ] = 2 s = 4 × 1 0 − 2 M ; pOH = 2 − 2 lo g 2 = 2 − 0.6 = 1.4 . pH = 14 − 1.4 = 12.6 . (Taking [ OH − ] = s would give 12.3.) Answer: (A).

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What is the pH of a saturated solution at $2 5^{\circ} C$ of slaked lime, $Ca (OH)_{2}$ , if its $K_{s p}$ is $3.2 \times 1 0^{- 5}$ ? [ $lo g 2 = 0.3$ ]

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Answer:

Solution:

$K_{s p} = 4 s^{3} = 3.2 \times 1 0^{- 5}$ , so $s^{3} = 8 \times 1 0^{- 6}$ and $s = 2 \times 1 0^{- 2} M$ .

$[OH^{-}] = 2 s = 4 \times 1 0^{- 2} M$ ; $pOH = 2 - 2 lo g 2 = 2 - 0.6 = 1.4$ .

$pH = 14 - 1.4 = 12.6$ .

(Taking $[OH^{-}] = s$ would give 12.3.)

Answer: (A).

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: QB

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