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What is the pH of a solution obtained by dissolving 0.0005 mole of the strong electrolyte, calcium hydroxide, $Ca {(OH)}_{2}$ to form 100 ml of a saturated solution (aqueous)? $(K_{W} = 1.0 \times 10^{- 14} {mole}^{2} {Liter}^{- 2})$

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12.0

3.0

9.8

11.7

answer is C.

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Detailed Solution

$M = \frac{n \times 1000}{V_{(mL)}}$

n = 0.0005

$Ca {(OH)}_{2};$ Electron. F = 2[Base's Acidity ]

$M = \frac{0.005 \times 1000}{100} = 0.005$

$N = n^{1} (M)$

$= 2 (0.005) = 10^{- 2} M$

$POH = - \log_{10}^{[OH]} = - \log_{10}^{10^{- 2}} = 2$

pH = pOH = 14

pH = 14 − 2 = 12

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