The equilibrium constant for the reactions
` Cu (s)+ 2Ag^(+) aq hArr Cu^(2+) (aq) +2Ag(s)`
is ` 2.0 xx 10 ^(15) " at " 278 K. ` Find the equilibrium concentration of `Cu^(2+) ` (aq) if that of `Ag^(+) ` (aq) is ` 1.0 xx 10 ^(-11) " mol " L^(-1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression
For the reaction:
\[ \text{Cu (s)} + 2 \text{Ag}^+ (aq) \rightleftharpoons \text{Cu}^{2+} (aq) + 2 \text{Ag (s)} \]

The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} \]
Note that the concentrations of solids (Cu and Ag) do not appear in the expression.

### Step 2: Substitute the known values
We know:
- \( K_c = 2.0 \times 10^{15} \)
- \( [\text{Ag}^+] = 1.0 \times 10^{-11} \, \text{mol L}^{-1} \)

Substituting these values into the equilibrium expression:
\[ 2.0 \times 10^{15} = \frac{[\text{Cu}^{2+}]}{(1.0 \times 10^{-11})^2} \]

### Step 3: Calculate \( (1.0 \times 10^{-11})^2 \)
Calculating the square of the concentration of \( \text{Ag}^+ \):
\[ (1.0 \times 10^{-11})^2 = 1.0 \times 10^{-22} \]

### Step 4: Rearrange the equation to solve for \( [\text{Cu}^{2+}] \)
Now we can rearrange the equation to solve for \( [\text{Cu}^{2+}] \):
\[ [\text{Cu}^{2+}] = K_c \times (1.0 \times 10^{-11})^2 \]
\[ [\text{Cu}^{2+}] = 2.0 \times 10^{15} \times 1.0 \times 10^{-22} \]

### Step 5: Perform the multiplication
Calculating the right side:
\[ [\text{Cu}^{2+}] = 2.0 \times 10^{-7} \, \text{mol L}^{-1} \]

### Final Answer
The equilibrium concentration of \( \text{Cu}^{2+} \) is:
\[ [\text{Cu}^{2+}] = 2.0 \times 10^{-7} \, \text{mol L}^{-1} \]

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