$K_{a}$ for 0.1 M $C H_{3} C O O H$ will be if it dissociate by 1.30% :-

1.69 \times 10^{- 5}

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1.69 \times 10^{- 4}

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1.69 \times 10^{- 6}

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None of these

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Solution

The correct option is C $1.69 \times 10^{- 5}$
$C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}$

$c$ $-$ $-$

$c (1 - x)$ $c x$ $c x$

$K_{a} = \frac{c x \times c x}{c (1 - x)} = c x^{2} = 0.1 \times (0.013)^{2} =$ $1.69 \times 10^{- 5}$

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The ionization energy of hydrogen atom is -13.6 eV. The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is: (Avogadro's constant =6.022 × 10²³)

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Ka for 0.1 M CH3COOH will be if it dissociate by 1.30% :-

The correct option is C 1.69×10−5CH3COOH→CH3COO−+H+ c − − c(1−x) cx cxKa=cx×cxc(1−x)=cx2=0.1×(0.013)2= 1.69×10−5

$K_{a}$ for 0.1 M $C H_{3} C O O H$ will be if it dissociate by 1.30% :-

The correct option is C $1.69 \times 10^{- 5}$
$C H_{3} C O O H \to C H_{3} C O O^{-} + H^{+}$

$c$ $-$ $-$

$c (1 - x)$ $c x$ $c x$

$K_{a} = \frac{c x \times c x}{c (1 - x)} = c x^{2} = 0.1 \times (0.013)^{2} =$ $1.69 \times 10^{- 5}$