Calculate K for the reaction, `A^(-)+H_(3)^(+)O hArr HA+H_(2)O`
if `K_(a)` value for the acid HA is `1.0xx10^(-6)`.

Options

A. `1.0 xx 10^(6)`
B. `1.0 xx 10^(-8)`
C. `1.0 xx 10^(8)`
D. `1.0 xx 10^(-6)`

Explanation

To calculate the equilibrium constant \( K \) for the reaction

\[ A^- + H_3O^+ \rightleftharpoons HA + H_2O \]

given that the \( K_a \) value for the acid \( HA \) is \( 1.0 \times 10^{-6} \), we can follow these steps:

### Step 1: Write the dissociation reaction of the acid \( HA \)

The dissociation of the acid \( HA \) in water can be represented as:

\[ HA + H_2O \rightleftharpoons H_3O^+ + A^- \]

### Step 2: Identify the equilibrium constant for the dissociation reaction

The equilibrium constant \( K_a \) for this dissociation reaction is given as:

\[ K_a = \frac{[H_3O^+][A^-]}{[HA][H_2O]} \]

Since the concentration of water \( [H_2O] \) is constant in dilute solutions, we can simplify this to:

\[ K_a = \frac{[H_3O^+][A^-]}{[HA]} \]

### Step 3: Reverse the dissociation reaction

To find the equilibrium constant for the reaction we are interested in, we need to reverse the dissociation reaction:

\[ H_3O^+ + A^- \rightleftharpoons HA + H_2O \]

### Step 4: Relate the equilibrium constants

When a reaction is reversed, the equilibrium constant for the new reaction is the reciprocal of the equilibrium constant of the original reaction. Therefore, we have:

\[ K = \frac{1}{K_a} \]

### Step 5: Substitute the value of \( K_a \)

Substituting the given \( K_a \) value into the equation:

\[ K = \frac{1}{1.0 \times 10^{-6}} = 1.0 \times 10^{6} \]

### Final Answer

Thus, the equilibrium constant \( K \) for the reaction

\[ A^- + H_3O^+ \rightleftharpoons HA + H_2O \]

is

\[ K = 1.0 \times 10^{6} \]

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