What is the hydronium ion concentration of a `0.02` M solution of `Cu^(2+)` solution of copper(II) perchlorate? The acidity constant of the following reaction is `5xx10^(-9)`.
`Cu^(2+)(aq.)+2H_(2)O(l)hArrCu(OH)^(+)(aq.)+H_(3)O^(+)(aq.)`
(a)`1xx10^(-5)`
(b)`7xx10^(-4)`
(c)`5xx10^(-4)`
(d)`1xx10^(-4)`

Options

A. `1xx10^(-5)`
B. `7xx10^(-4)`
C. `5xx10^(-4)`
D. `1xx10^(-4)`

Answer

To find the hydronium ion concentration of a `0.02 M` solution of `Cu^(2+)` from the given reaction and acidity constant, we can follow these steps:

### Step 1: Write the Reaction
The reaction given is:
\[ \text{Cu}^{2+}(aq) + 2 \text{H}_2\text{O}(l) \rightleftharpoons \text{CuOH}^+(aq) + \text{H}_3\text{O}^+(aq) \]

### Step 2: Identify Initial Concentrations
The initial concentration of `Cu^(2+)` is `0.02 M`. We can denote this concentration as \( C = 0.02 \, \text{M} \).

### Step 3: Set Up the Equilibrium Expression
Let \( \alpha \) be the degree of dissociation of `Cu^(2+)`. At equilibrium, the concentrations will be:
- For `Cu^(2+)`: \( C(1 - \alpha) \)
- For `CuOH^+`: \( C\alpha \)
- For `H3O^+`: \( C\alpha \)

The equilibrium expression for the acidity constant \( K_a \) is given by:
\[ K_a = \frac{[\text{CuOH}^+][\text{H}_3\text{O}^+]}{[\text{Cu}^{2+}]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 4: Simplify the Expression
Assuming \( \alpha \) is very small compared to 1 (which is valid for weak acids), we can simplify the expression:
\[ K_a \approx \frac{C\alpha^2}{C} = C\alpha^2 \]

### Step 5: Substitute Known Values
We know:
- \( K_a = 5 \times 10^{-9} \)
- \( C = 0.02 \)

Substituting these values into the equation:
\[ 5 \times 10^{-9} = 0.02 \alpha^2 \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha^2 = \frac{5 \times 10^{-9}}{0.02} \]
\[ \alpha^2 = 2.5 \times 10^{-7} \]
\[ \alpha = \sqrt{2.5 \times 10^{-7}} = 5 \times 10^{-4} \]

### Step 7: Calculate Hydronium Ion Concentration
The concentration of hydronium ions \([H_3O^+]\) is given by:
\[ [H_3O^+] = C\alpha = 0.02 \times 5 \times 10^{-4} \]
\[ [H_3O^+] = 1 \times 10^{-5} \, \text{M} \]

### Step 8: Conclusion
Thus, the hydronium ion concentration of the solution is:
\[ \boxed{1 \times 10^{-5} \, \text{M}} \]