The number of moles of NH3, that must be added to 2 L of 0.80 M AgNO3 in order to reduce the concent

Q: Why do we assume the concentration of the complex is exactly $0.80\text{ M}$?

Because the formation constant ($K_f = 10^8$) is extremely large. This means the equilibrium lies far to the right. Almost all of the initial $0.80\text{ M}$ $\text{Ag}^+$ is converted into the complex. The tiny amount of $\text{Ag}^+$ left ($5.0 \times 10^{-8}\text{ M}$) is negligible compared to $0.80\text{ M}$.

Q: Why is the equilibrium concentration of $\text{NH}_3$ written as $\frac{a}{2} - 1.6$?

We started with $a$ moles of $\text{NH}_3$ in $2\text{ L}$, so its initial concentration is $a/2\text{ M}$. The stoichiometry shows that 2 moles of $\text{NH}_3$ react for every 1 mole of $\text{Ag}^+$. Since $0.80\text{ M}$ of $\text{Ag}^+$ reacted, $2 \times 0.80 = 1.6\text{ M}$ of $\text{NH}_3$ must have been consumed.

Q: Can I solve this using moles instead of molarity?

Yes, absolutely! You can set up the ICE table using moles and then divide by the total volume ($2\text{ L}$) when plugging into the $K_f$ expression. Both methods will yield the exact same result.

Q: What if the volume changed upon adding $\text{NH}_3$?

If the volume changed, we would need to know the final volume to calculate the new concentrations. The problem explicitly states to assume no volume change to simplify the calculation.

Q: Why did we take the positive square root of 0.16?

The concentration of a chemical species cannot be negative. Therefore, $\frac{a}{2} - 1.6$ must be a positive value, so we take the principal square root, which is $0.4$.

Question

The number of moles of $	ext{NH}_3$, that must be added to $2	ext{ L}$ of $0.80	ext{ M}$ $	ext{AgNO}_3$ in order to reduce the concentration of $	ext{Ag}^+$ ions to $5.0 	imes 10^{-8}	ext{ M}$ ($K_{	ext{formation}}$ for $[	ext{Ag}(	ext{NH}_3)_2]^+ = 1.0 	imes 10^8$) is ......... (Nearest integer)
[Assume no volume change on adding $	ext{NH}_3$]

Sigma Percentile · Accepted Answer

### The Magic of Complexation

Imagine trying to hide a needle in a haystack. That is exactly what we are doing with the silver ions in this problem. We are using ammonia to "hide" them inside a complex ion.

This is a classic application of chemical equilibrium where the formation constant is incredibly large.

### Analyzing the Setup

We start with a beaker containing $2	ext{ L}$ of a $0.80	ext{ M}$ $	ext{AgNO}_3$ solution. This means our initial concentration of silver ions, $[	ext{Ag}^+]$, is $0.80	ext{ M}$.

We are tasked with finding the number of moles of ammonia, let's call it $a$, that we need to add to drastically reduce the free silver ion concentration.

Since the volume remains constant at $2	ext{ L}$, the initial concentration of the added ammonia will simply be $\frac{a}{2}	ext{ M}$.

### The Master Equation

When ammonia is introduced, it reacts with the silver ions to form the diamminesilver(I) complex. The balanced chemical equation is:

$$ 	ext{Ag}^+ + 2	ext{NH}_3 ightleftharpoons [	ext{Ag}(	ext{NH}_3)_2]^+ $$

Notice the stoichiometry carefully. One mole of silver ions requires exactly two moles of ammonia to form the complex.

The equilibrium is governed by the formation constant, $K_f$, which is given as $1.0 	imes 10^8$.

$$ K_f = \frac{[	ext{Ag}(	ext{NH}_3)_2^+]}{[	ext{Ag}^+][	ext{NH}_3]^2} $$

### The "Completion" Approximation

Here is where the magic happens. The value of $K_f$ is $10^8$, which is massive! This tells us that the forward reaction is highly favored.

For all practical purposes, the reaction goes almost to completion. Nearly all of the initial $0.80	ext{ M}$ silver ions will be converted into the complex.

Therefore, the equilibrium concentration of the complex, $[	ext{Ag}(	ext{NH}_3)_2^+]$, will be approximately $0.80	ext{ M}$.

What about the ammonia? Since $0.80	ext{ M}$ of silver reacted, it must have consumed twice that amount of ammonia, which is $1.6	ext{ M}$.

So, the equilibrium concentration of ammonia will be its initial concentration minus what was consumed: $\left(\frac{a}{2} - 1.6ight)	ext{ M}$.

We are also given that the final, tiny concentration of free silver ions is $5.0 	imes 10^{-8}	ext{ M}$.

### Final Calculation

Now, we substitute all these equilibrium values into our $K_f$ expression.

$$ 1.0 	imes 10^8 = \frac{0.80}{(5.0 	imes 10^{-8}) \left(\frac{a}{2} - 1.6ight)^2} $$

At first glance, this might look like a messy calculation. But look closely at the powers of ten!

When we rearrange the equation, the $10^8$ and $10^{-8}$ perfectly cancel each other out.

$$ \left(\frac{a}{2} - 1.6ight)^2 = \frac{0.80}{10^8 	imes 5.0 	imes 10^{-8}} $$

$$ \left(\frac{a}{2} - 1.6ight)^2 = \frac{0.80}{5.0} = 0.16 $$

Taking the square root of both sides gives us a beautifully simple linear equation.

$$ \frac{a}{2} - 1.6 = 0.4 $$

Solving for $a$, we get $\frac{a}{2} = 2.0$, which means **$a = 4$**.

We must add exactly **4 moles** of ammonia to achieve the desired reduction in silver ion concentration.

Enter Numerical Value:

Visualized Solution

\text{Initial State}

\text{Complex Formation Reaction}

\text{Initial Concentrations}

\text{Equilibrium Concentrations}

\text{Equilibrium Constant Expression}

\text{Substituting Values}

\text{Solving for Ammonia Concentration}

\text{Final Calculation}

The Sigma Insight: Law of Mass Action

The Magic of Complexation

Analyzing the Setup

The Master Equation

The "Completion" Approximation

Final Calculation

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