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For a weak acid HA of concentration C(mo...

For a weak acid HA of concentration `C(mol l^-1)` and degree of dissociation `(alpha)`, Ostwald's dilution law is represented by the equation

A

`K_a=(C^2alpha)/(1-alpha)`

B

`K_a=(alpha^2C)/(1-alpha)`

C

`K_a= Calpha`

D

`K_a= (Calpha^2)/(1-alpha^2)`

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The correct Answer is:
To solve the problem regarding Ostwald's dilution law for a weak acid HA, we will follow these steps: ### Step 1: Understand the dissociation of the weak acid The weak acid HA dissociates in water according to the following equilibrium reaction: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] ### Step 2: Define initial concentration and degree of dissociation Let the initial concentration of the weak acid HA be \( C \) (in mol/L). The degree of dissociation is denoted by \( \alpha \), which represents the fraction of the acid that has dissociated. ### Step 3: Set up equilibrium concentrations At equilibrium: - The concentration of undissociated acid (HA) will be \( C(1 - \alpha) \). - The concentration of \( \text{H}^+ \) ions will be \( C\alpha \). - The concentration of \( \text{A}^- \) ions will also be \( C\alpha \). ### Step 4: Write the expression for the equilibrium constant \( K_a \) The equilibrium constant \( K_a \) for the dissociation of the weak acid can be expressed as: \[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \] Substituting the equilibrium concentrations into this expression gives: \[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \] ### Step 5: Simplify the equation This simplifies to: \[ K_a = \frac{C^2 \alpha^2}{C(1 - \alpha)} \] Cancelling one \( C \) from the numerator and denominator results in: \[ K_a = \frac{C \alpha^2}{1 - \alpha} \] ### Step 6: Identify the correct option From the derived equation, we can see that the correct representation of Ostwald's dilution law for the weak acid HA is: \[ K_a = \frac{C \alpha^2}{1 - \alpha} \] This corresponds to option 2 in the provided choices. ### Final Answer The correct equation representing Ostwald's dilution law for the weak acid HA is: \[ K_a = \frac{C \alpha^2}{1 - \alpha} \] ---
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AAKASH INSTITUTE ENGLISH-MOCK TEST 13-Exercise
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  2. An example of a strong electrolyte is

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  3. For a weak acid HA of concentration C(mol l^-1) and degree of dissocia...

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  5. pH of a 0.001 M NaOH solution will be

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  7. The dissociation constant of an acid, HA is 1 x 10^-5 The pH of 0.1 M ...

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  8. 100 mL of 0.01 M solution of NaOH is diluted to 1 litre. The pH of res...

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  9. At 80 ^(@ ) C , pure distilled water has [H3O^(+) ]= 1 xx 10 ^(-6) ...

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  10. The pH of a solution obtained by mixing 50 mL of 2N HCI and 50 mL of 1...

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  11. The pH of a solution increased from 3 to 6. Its [H^(o+)] will be

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  12. Ionic product of water increases, if

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  14. A monobasic weak acid solution which is 0.002 M has pH value equal to ...

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  17. Aqueous solution of sodium acetate is

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  18. An acidic buffer solution can be prepared by mixing the solutions of

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  19. Degree hydrolysis (h) of a salt of weak acid and a strong base is give...

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  20. The pH of a solution at 25°C containing 0.20 M sodium acetate and 0.06...

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