The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of $3 %$ in its $0.1 M$ solution at $25^{0} C$ . Given that the ionic product of water is $10^{- 14}$ at this temperature, what is the dissociation constant of the acid?

\approx 1 \times 10^{- 10}

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\approx 1 \times 10^{- 9}

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3.33 \times 10^{- 9}

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3.33 \times 10^{- 10}

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Solution

The correct option is C $\approx 1 \times 10^{- 10}$
Let weak acid is $H A$ and its sodium salt is $N a A$ .
$K_{a} = \frac{K_{W}}{K_{H}}$ ; $K_{H} = C H^{2} \Rightarrow 0.1 \times {(0.03)}^{2}$

$K_{a} = \frac{10^{- 14}}{9 \times 10^{- 5}}$ ; $K_{a} \approx 1 \times 10^{- 10}$

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The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of 3% in its 0.1 M solution at 250C. Given that the ionic product of water is 10−14 at this temperature, what is the dissociation constant of the acid?

The correct option is C ≈1×10−10Let weak acid is HA and its sodium salt is NaA.Ka=KWKH ; KH=CH2⇒0.1×(0.03)2Ka=10−149×10−5 ; Ka≈1×10−10

The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of $3 %$ in its $0.1 M$ solution at $25^{0} C$ . Given that the ionic product of water is $10^{- 14}$ at this temperature, what is the dissociation constant of the acid?

The correct option is C $\approx 1 \times 10^{- 10}$
Let weak acid is $H A$ and its sodium salt is $N a A$ .
$K_{a} = \frac{K_{W}}{K_{H}}$ ; $K_{H} = C H^{2} \Rightarrow 0.1 \times {(0.03)}^{2}$

$K_{a} = \frac{10^{- 14}}{9 \times 10^{- 5}}$ ; $K_{a} \approx 1 \times 10^{- 10}$