SATHEE: Equilibrium Question 901

Equilibrium Question 901

Question: A solution saturated in lime water has a pH of 12.4. Then the Ksp for $C a {(O H)}_{2}$ is:

Options:

A) $3.2 \times 10^{- 3}$

B) $7.8 \times 10^{- 6}$

C) $7.8 \times 10^{- 28}$

D) $3.2 \times 10^{- 4}$

Show Answer

Answer:

Correct Answer: B

Solution:

$p H = 12.4$

$\Rightarrow$ $- \log (H^{+}) = 12.4, \Rightarrow$

$\log [H^{+}] = 13.6$

$C a {(\overset{+}{O H},)}_{2} \to 2 O H + C a^{2 +}$

$[H^{+}] = 4 \times 10^{- 13}$

$[O H^{-}] = \frac{10^{- 14}}{4 \times 10} = \frac{1}{3} \times 10^{- 1} = 2.5 \times 10^{- 2}$

$K s p = [C a^{2 +}] [O H^{-}]$

$[C a^{2 +}] = \frac{1}{2} [O H^{-}]$

$(\frac{1}{8} \times 10^{- 1}) {(\frac{1}{4} \times 10^{- 1})}^{2} = \frac{1}{8} \times \frac{1}{16} \times 10^{- 3}$

$7.8 \times 10^{- 6}$

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Equilibrium Question 901

Question: A solution saturated in lime water has a pH of 12.4. Then the Ksp for Ca(OH)2 is:

Options:

Answer:

Solution:

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Question: A solution saturated in lime water has a pH of 12.4. Then the Ksp for $C a {(O H)}_{2}$ is: