The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)`
`H_(3)ClO_(4)` is a tribasic acid, it undergoes ionisation as
`H_(3)ClO_(4) hArr H^(o+)+H_(2)ClO_(4)^(-), K_(1)`
`H_(2)ClO_(4)^(-) hArr H^(o+)+HClO_(4)^(2-), K_(2)`
`HClO_(4)^(2-) hArr H^(o+)+ClO_(4)^(3-), K_(3)`
Then, equilibrium constant for the following reaction will be:
`H_(3)ClO_(4) hArr 3H^(o+)+ClO_(4)^(3-)`

Options

A. `K_(1)K_(2)K_(3)`
B. `((K_(1)K_(3))^(2))/(K_(2))`
C. `K_(1)/K_(2)`
D. `(K_(1)K_(2))/(K_(3)^(2))`

Explanation

To find the equilibrium constant for the reaction \( H_3ClO_4 \rightleftharpoons 3H^+ + ClO_4^{3-} \), we will analyze the ionization steps of the tribasic acid \( H_3ClO_4 \) and relate them to the final reaction.

### Step 1: Write down the ionization reactions and their equilibrium constants.

1. The first ionization reaction:
\[ H_3ClO_4 \rightleftharpoons H^+ + H_2ClO_4^- \]
with equilibrium constant \( K_1 \).

2. The second ionization reaction:
\[ H_2ClO_4^- \rightleftharpoons H^+ + HClO_4^{2-} \]
with equilibrium constant \( K_2 \).

3. The third ionization reaction:
\[ HClO_4^{2-} \rightleftharpoons H^+ + ClO_4^{3-} \]
with equilibrium constant \( K_3 \).

### Step 2: Combine the ionization reactions.

To obtain the overall reaction:
\[ H_3ClO_4 \rightleftharpoons 3H^+ + ClO_4^{3-} \]
we can add the three ionization reactions together.

### Step 3: Add the equilibrium constants.

When we add the reactions, the equilibrium constant for the overall reaction \( K \) is the product of the individual equilibrium constants:
\[ K = K_1 \times K_2 \times K_3 \]

### Step 4: Conclusion.

Thus, the equilibrium constant for the reaction \( H_3ClO_4 \rightleftharpoons 3H^+ + ClO_4^{3-} \) is given by:
\[ K = K_1 \times K_2 \times K_3 \]

### Final Answer:
The equilibrium constant for the reaction \( H_3ClO_4 \rightleftharpoons 3H^+ + ClO_4^{3-} \) is \( K_1 \times K_2 \times K_3 \).

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