For dissociation constant (K) and ionic product $(K_w)$ of water which is correct.
A) $K > K_w$
B) $K_w > K$
C) $K_w = K$
D) None of these
Answer
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Hint: The dissociation constant (K) is a specific type of equilibrium constant that measures the property of a larger object of dissociating reversibly into smaller components.
Where Ionic product means, The product of their concentration of ions.
${H_2}O$ gets dissociated in ${H_3}{O^ + }$ and $O{H^ - }$ so the product of their concertation is an ionic product.
Complete step by step answer:
Water is a poor conductor of electricity.
\[{H_2}O\underset {} \leftrightarrows {H^ + } + O{H^ - }\]
and,
\[2{H_2}O\overset {} \leftrightarrows {H_3}O + O{H^ - }\]
Ionic product of water may be defined as product of $\left[ {{H^ + }} \right]$ and $\left[ {O{H^ - }} \right]$ where $[\;]$ denotes the concentration.
And also, can say that ${H^ + }$ ions in water exist as ${H_3}{O^ + }$ ions, therefore, ionic product of molar concentration of ${H_3}{O^ + }{\text{ and }}O{H^ - }$
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] \\
{K_w} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]..............(1) \\$
Dissociation constant of water is,
$K = \dfrac{{\left[ {{H^ + }} \right].\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}}$
And for dissociation of ${H_3}{O^ + }$ and $O{H^ - }$, we can say,
$K = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {2.{H_2}O} \right]}} = \dfrac{{{K_w}}}{{\left[ {2.{H_2}O} \right]}} = \dfrac{{{K_w}}}{{{{\left[ {{H_2}O} \right]}^2}}}$
[Put the value of ${K_w}$ from (1)].
Here, we can see, that dissociation constant is smaller too ${K_w}$ because ${\left[ {{H_2}O} \right]^2}$ is divided with ${K_w}$ to get, K.
$\therefore K < {K_w}$
$\therefore $ Option B is correct.
Note: We get by experiment that dissociation constant of water is $1.0 \times {10^{ - 7}}$. Water dissociates only slightly, as water is a weak electrolyte. Ionic product is the normal product of concentrate of ions, each raised to the power specified by its stoichiometric coefficient in a solution of a salt.
Where Ionic product means, The product of their concentration of ions.
${H_2}O$ gets dissociated in ${H_3}{O^ + }$ and $O{H^ - }$ so the product of their concertation is an ionic product.
Complete step by step answer:
Water is a poor conductor of electricity.
\[{H_2}O\underset {} \leftrightarrows {H^ + } + O{H^ - }\]
and,
\[2{H_2}O\overset {} \leftrightarrows {H_3}O + O{H^ - }\]
Ionic product of water may be defined as product of $\left[ {{H^ + }} \right]$ and $\left[ {O{H^ - }} \right]$ where $[\;]$ denotes the concentration.
And also, can say that ${H^ + }$ ions in water exist as ${H_3}{O^ + }$ ions, therefore, ionic product of molar concentration of ${H_3}{O^ + }{\text{ and }}O{H^ - }$
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] \\
{K_w} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]..............(1) \\$
Dissociation constant of water is,
$K = \dfrac{{\left[ {{H^ + }} \right].\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}}$
And for dissociation of ${H_3}{O^ + }$ and $O{H^ - }$, we can say,
$K = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {2.{H_2}O} \right]}} = \dfrac{{{K_w}}}{{\left[ {2.{H_2}O} \right]}} = \dfrac{{{K_w}}}{{{{\left[ {{H_2}O} \right]}^2}}}$
[Put the value of ${K_w}$ from (1)].
Here, we can see, that dissociation constant is smaller too ${K_w}$ because ${\left[ {{H_2}O} \right]^2}$ is divided with ${K_w}$ to get, K.
$\therefore K < {K_w}$
$\therefore $ Option B is correct.
Note: We get by experiment that dissociation constant of water is $1.0 \times {10^{ - 7}}$. Water dissociates only slightly, as water is a weak electrolyte. Ionic product is the normal product of concentrate of ions, each raised to the power specified by its stoichiometric coefficient in a solution of a salt.
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