20. What is the maximum concentration of Mg^2+ ion that remains dissolved in a solution that contains 0,7147 M NH_(3) and 0,2073MNH_(4)Cl (K_(sp) for Mg(OH)_(2) is 1,2times 10^-11;K_(b) for NH_(3) is 1,77times 10^-5 A. 1,9times 10^-7M B. 3,2times 10^-3M C. 1,3times 10^-3M D.. 6,4times 10^-6M

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Pertanyaan

20. What is the maximum concentration of Mg^2+ ion that remains dissolved in a solution that contains 0,7147 M NH_(3) and 0,2073MNH_(4)Cl (K_(sp) for Mg(OH)_(2) is 1,2times 10^-11;K_(b) for NH_(3) is 1,77times 10^-5 A. 1,9times 10^-7M B. 3,2times 10^-3M C. 1,3times 10^-3M D.. 6,4times 10^-6M

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Answer 1

To find the maximum concentration of \( \text{Mg}^{2+} \) that remains dissolved, we need to consider the solubility product constant (\( K_{sp} \)) for \( \text{Mg(OH)}_2 \) and the equilibrium expressions involving ammonia (\( \text{NH}_3 \)) and ammonium chloride (\( \text{NH}_4\text{Cl} \)).

First, write the \( K_{sp} \) expression for \( \text{Mg(OH)}_2 \):
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]

Given:
\[ K_{sp} = 1.2 \times 10^{-11} \]

Next, determine the concentration of hydroxide ions (\( \text{OH}^- \)) from the given concentrations of \( \text{NH}_3 \) and \( \text{NH}_4\text{Cl} \).

The equilibrium expression for \( \text{NH}_3 \) in water is:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

Using the base dissociation constant (\( K_b \)) for \( \text{NH}_3 \):
\[ K_b = 1.77 \times 10^{-5} \]

Set up the expression for \( K_b \):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

Assume \( x \) is the concentration of \( \text{OH}^- \) produced:
\[ K_b = \frac{x^2}{0.7147 - x} \]

Since \( x \) is typically very small compared to the initial concentration of \( \text{NH}_3 \), we can approximate:
\[ 1.77 \times 10^{-5} = \frac{x^2}{0.7147} \]

Solve for \( x \):
\[ x^2 = 1.77 \times 10^{-5} \times 0.7147 \]
\[ x^2 = 1.265 \times 10^{-5} \]
\[ x = \sqrt{1.265 \times 10^{-5}} \]
\[ x = 3.56 \times 10^{-3} \]

This \( x \) is the concentration of \( \text{OH}^- \).

Now, use the \( K_{sp} \) expression for \( \text{Mg(OH)}_2 \):
\[ 1.2 \times 10^{-11} = [\text{Mg}^{2+}](3.56 \times 10^{-3})^2 \]

Solve for \( [\text{Mg}^{2+}] \):
\[ [\text{Mg}^{2+}] = \frac{1.2 \times 10^{-11}}{(3.56 \times 10^{-3})^2} \]
\[ [\text{Mg}^{2+}] = \frac{1.2 \times 10^{-11}}{1.265 \times 10^{-5}} \]
\[ [\text{Mg}^{2+}] = 9.46 \times 10^{-7} \]

Therefore, the maximum concentration of \( \text{Mg}^{2+} \) that remains dissolved is approximately \( 9.46 \times 10^{-7} \) M, which is closest to option A.

Answer: A. \( 1.9 \times 10^{-7} \) M

Pertanyaan

20. What is the maximum concentration of Mg^2+ ion that remains dissolved in a solution that contains 0,7147 M NH_(3) and 0,2073MNH_(4)Cl (K_(sp) for Mg(OH)_(2) is 1,2times 10^-11;K_(b) for NH_(3) is 1,77times 10^-5 A. 1,9times 10^-7M B. 3,2times 10^-3M C. 1,3times 10^-3M D.. 6,4times 10^-6M

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