Question

A solution is made by mixing 500.0 mL of 4.0 M NH3 and 500.0 mL of 0.50 M AgNO3. Ag+ reacts with NH3 to form AgNH3+ and Ag(NH3)2+ according to the equilibrium reactions:

Ag+ + NH3 ↔ AgNH3+ K¬1 = 2.1 x 103

AgNH3+ + NH3 ↔ Ag(NH3)2+ K¬2 = 8.2 x 103

Assuming no change in volume on mixing calculate the concentrations of all species in solution: [Ag+], [NO3-], [NH3], [AgNH3+] and [Ag(NH3)2+].

Answer 1

Answer on Question #56945 - Chemistry - General Chemistry

Question:

A solution is made by mixing $500.0~\mathrm{mL}$ of $4.0\mathrm{M}$ NH3 and $500.0~\mathrm{mL}$ of $0.50\mathrm{M}$ AgNO3. $\mathrm{Ag^{+}}$ reacts with NH3 to form $\mathrm{AgNH3 + }$ and $\mathrm{Ag(NH3)}2+$ according to the equilibrium reactions:

$\mathrm{Ag + + NH3}\leftrightarrow \mathrm{AgNH3 + K\neg 1 = 2.1\times 103}$

$\mathrm{AgNH3 + + NH3}\leftrightarrow \mathrm{Ag(NH3)}2 + \mathrm{K}\neg 2 = 8.2\times 103$

Assuming no change in volume on mixing calculate the concentrations of all species in solution: $[\mathrm{Ag^{+}}]$ , [NO3-], [NH3], $[\mathrm{AgNH3^{+}}]$ and $[\mathrm{Ag(NH3)}2^{+}]$ .

Answer:

After mixing the initial concentrations of $\mathrm{AgNO}_3$ and $\mathrm{NH}_3$ are of 0.5 and 4 mol/L. Since $\mathrm{AgNO}_3$ dissociates completely, the concentration of $\mathrm{NO}_3^-$ is 0.5 mol/L too.

The equilibrium constants are connected with concentrations of the components as follows:

$\mathrm{K}_{1} = [\mathrm{AgNH}_{3}^{+}] / ([\mathrm{Ag}^{+}][\mathrm{NH}_{3}])$

$\mathrm{K}_{2} = [\mathrm{Ag}(\mathrm{NH}_{3})_{2}^{+}] / ([\mathrm{AgNH}_{3}^{+}][\mathrm{NH}_{3}])$

The common equilibrium constant for the formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$ :

$\mathrm{Ag}^{+} + 2 \mathrm{NH}_{3} \rightarrow \left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$

$\mathrm{K} = \mathrm{K}_{1}\mathrm{K}_{2} = [\mathrm{Ag}(\mathrm{NH}_{3})_{2}^{+}] / ([\mathrm{Ag}^{+}][\mathrm{NH}_{3}]^{2}) = 1.722\times 10^{7}$

Assuming that $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] = z$ , we get $\left[\mathrm{Ag}^{+}\right] = 0.5 - z$ and $\left[\mathrm{NH}_{3}\right] = 4 - 2z$ .

Substituting the parameters into the equilibrium constant expression, it is obtained:

$\mathrm{K} = \mathrm{z} / \left[ (0.5 - \mathrm{z})(4 - 2\mathrm{z})^{2} \right] = 1.722 \times 10^{7}$

$z = 4K(0.5 - z)(2 - z)^{2}$

$z = 4K(0.5 - z)(4 - 4z + z^{2})$

$z = 4K(2 - 2z + 0.5z^{2} - 4z + 4z^{2} - z^{3})$

$z = 8K - 24Kz + 18Kz^{2} - 4Kz^{3}$

$2\mathrm{Kz}^{3} - 9\mathrm{Kz}^{2} + (12\mathrm{K} + 0.5)\mathrm{z} - 4\mathrm{K} = 0$

$$3.444z^3 - 15.498z^2 + 20.664z - 6.888 = 0$$

The solution of the cubic equation gives:

$$z_1 = 0.4999999999999998$$

Thus, $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2^*\right] = z = 0.4999999999999998\ \mathrm{mol/L}$

then

$$\left[\mathrm{Ag}^+\right] = 0.5 - 0.4999999999999998 = 2 \times 10^{-16}\ \mathrm{mol/L}$$

and

$$\left[\mathrm{NH}_3\right] = 4 - 2z \approx 3\ \mathrm{mol/L}$$

Using the expression for $K_1$ the concentration of $\mathrm{AgNH}_3^+$ is found:

$$\left[\mathrm{AgNH}_3^+\right] = K_1 \left[\mathrm{Ag}^+\right] \left[\mathrm{NH}_3\right] = 2.1 \times 10^3 \times 2 \times 10^{-16} \times 3 = 1.26 \times 10^{-12}\ \mathrm{mol/L}$$

Finally, the concentrations of all components are:

$$\left[\mathrm{NO}_3^-\right] = 0.5\ \mathrm{mol/L}$$ $$\left[\mathrm{Ag}^+\right] = 2 \times 10^{-16}\ \mathrm{mol/L}$$ $$\left[\mathrm{NH}_3\right] \approx 3\ \mathrm{mol/L}$$ $$\left[\mathrm{AgNH}_3^+\right] = 1.26 \times 10^{-12}\ \mathrm{mol/L}$$ $$\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2^*\right] = 0.4999999999999998\ \mathrm{mol/L}$$

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