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Solubility Product

Which of the ...

Question

Which of the following concentrations of $N H_{4}^{+}$ will be sufficient to prevent the precipitation of $M g (O H)_{2}$ from a solution which is $0.01 {M MgCl}_{2}$ and $0.1 {M NH}_{3} (a q)$ ? Given that the $K_{s p}$ of $M g (O H)_{2} = 2.5 \times 10^{- 11}$ and $K_{b}$ for $N H_{3} (a q) = 2 \times 10^{- 5}$ .

A

0.01 M

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B

0.02 M

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C

0.03 M

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D

0.04 M

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Solution

The correct option is D 0.04 M
$K_{s p} M g (O H)_{2} = [M g^{2 +}] [O H^{-}]^{2} = 2.5 \times 10^{- 11}$
$∴ [O H^{-}] = \sqrt{\frac{2.5 \times 10^{- 11}}{0.01}} = 5 \times 10^{- 5}$
now $K_{b} N H_{3} = 2 \times 10^{- 5} = \frac{[N H_{4}^{+}] [O H^{-}]}{[N H_{3}]}$
hence
$[N H_{4}^{+}] = 0.04 M$

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Similar questions

Q. Which of the following concentrations of

N H_{4}^{+}

will be sufficient to prevent the precipitation of

M g (O H)_{2}

from a solution which is

0.01 {M MgCl}_{2}

0.1 {M NH}_{3} (a q)

? Given that the

K_{s p}

M g (O H)_{2} = 2.5 \times 10^{- 11}

K_{b}

N H_{3} (a q) = 2 \times 10^{- 5}

.

Q. Which of the following concentrations of

N H_{4}^{+}

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M g (O H)_{2}

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K_{s p}

M g (O H)_{2} = 2.5 \times 10^{- 11}

K_{b}

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Q. Which of the following concentrations of

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will be sufficient to prevent the precipitation of

M g (O H)_{2}

form a solution which is 0.01 M

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M g (O H)_{2} = 2.5 \times 10^{- 11}

K_{b}

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Q. Which of the following concentrations of

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will be sufficient to prevent the precipitation of

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0.01 {M MgCl}_{2}

0.1 {M NH}_{3} (a q)

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M g (O H)_{2} = 2.5 \times 10^{- 11}

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Q. What is the maximum pH of a solution

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