What is the acidity constant for the following reaction given that the hydronium ion concentration of a 0.04 M solution of `Ni^(2+)` solution of nickel(II) perchlorate is `4.5xx10^(-6)`?
`Ni^(2+)(aq.)+2H_(2)O(l)hArrNi(OH)^(+)(aq.)+H_(3)O^(+)(aq.)`

Options

A. `2xx10^(-12)`
B. `4xx10^(-6)`
C. `5xx10^(-12)`
D. `5xx10^(-10)`

Answer

To find the acidity constant (Ka) for the reaction:

\[ \text{Ni}^{2+}(aq) + 2 \text{H}_2\text{O}(l) \rightleftharpoons \text{NiOH}^+(aq) + \text{H}_3\text{O}^+(aq) \]

we will follow these steps:

### Step 1: Write the expression for the acidity constant (Ka)
The acidity constant for the reaction can be expressed as:

\[ K_a = \frac{[\text{NiOH}^+][\text{H}_3\text{O}^+]}{[\text{Ni}^{2+}][\text{H}_2\text{O}]^2} \]

Since water is a pure liquid, its concentration remains constant and can be omitted from the expression. Thus, we have:

\[ K_a = \frac{[\text{NiOH}^+][\text{H}_3\text{O}^+]}{[\text{Ni}^{2+}]} \]

### Step 2: Set up the initial and equilibrium concentrations
Given:
- Initial concentration of \(\text{Ni}^{2+}\) = 0.04 M
- Hydronium ion concentration \([\text{H}_3\text{O}^+]\) = \(4.5 \times 10^{-6}\) M

At equilibrium, let \(x\) be the change in concentration:
- \([\text{Ni}^{2+}] = 0.04 - x\)
- \([\text{NiOH}^+] = x\)
- \([\text{H}_3\text{O}^+] = x\)

### Step 3: Substitute the known values
Since we know that \(x = [\text{H}_3\text{O}^+] = 4.5 \times 10^{-6}\), we can substitute:

\[ K_a = \frac{(4.5 \times 10^{-6})(4.5 \times 10^{-6})}{0.04 - 4.5 \times 10^{-6}} \]

### Step 4: Simplify the denominator
Since \(4.5 \times 10^{-6}\) is much smaller than 0.04, we can approximate:

\[ K_a \approx \frac{(4.5 \times 10^{-6})^2}{0.04} \]

### Step 5: Calculate \(K_a\)
Calculating \( (4.5 \times 10^{-6})^2 \):

\[ (4.5 \times 10^{-6})^2 = 20.25 \times 10^{-12} \]

Now substituting this value into the \(K_a\) expression:

\[ K_a \approx \frac{20.25 \times 10^{-12}}{0.04} \]

Calculating the division:

\[ K_a \approx 20.25 \times 10^{-12} \div 0.04 = 20.25 \div 4 \times 10^{-10} = 5.0625 \times 10^{-10} \]

### Step 6: Round to significant figures
Rounding gives us:

\[ K_a \approx 5 \times 10^{-10} \]

### Final Answer
Thus, the acidity constant \(K_a\) for the reaction is approximately:

\[ \boxed{5 \times 10^{-10}} \]