
Why is pH of pure water7?
(A) Concentration of hydroxide ion and hydrogen ion are equal.
(B) Concentration of hydroxide ion and hydrogen ion is zero.
(C) Concentration of hydroxide ion is more and hydrogen ion is less
(D) None of above
Answer
233.4k+ views
Hint: Try to recall that pH scale is used to determine the concentration of hydrogen ions in a solution. Since, we know the ionization product of ordinary water is $1 \times {10^{ - 14}}$ at 25 degree Celsius. Now by knowing the ionization product of water you can easily answer.
Complete step by step solution:
You should know that pH is defined as “the negative logarithm of the concentration of hydronium ion in a solution i.e. $pH = \log \left[ {{H^ + }} \right]$”.
It is known to you that water self-ionizes itself into hydronium $({H^ + })$ and hydroxide $(O{H^ - }\ ions$: ${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$ and the equilibrium constant for this reaction is called ionization constant
$({K_w})$.
Also, it is known to use from experiment that ionic product of water ,${K_w}$ at 298K is $1 \times {10^{ - 14}}$ and is expressed as ${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$.
Since, water is neutral so it should have equal concentration of both ${H^ + }$ and $O{H^ - }$ ions.
Now, Calculation:
Let concentration of ${H^ + }$ be x.
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]=1 \times {10^{ - 14}}$.
Since, from above it is pretty obvious that $\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]$,
\[
\Rightarrow x \times x = {10^{ - 14}}
\Rightarrow {x^2} = {10^{ - 14}}
\Rightarrow x = {10^{ - 7}}
\Rightarrow \left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]{10^{ - 7}}
\].
Therefore, from above calculations you can now clearly conclude that option A is the correct option to the given question.
Note: It should be remembered that the ionic product of water, ${K_w}$ increases with increase in temperature. This happens because of the fact that with increase, the degree of ionization of water increases i.e. more of molecules dissociate into ${H^ + }$ and $O{H^ - }$ ions. Thus, concentration of ${H^ + }$ and $O{H^ - }$ ions increases and hence, ionic products also increases.
Complete step by step solution:
You should know that pH is defined as “the negative logarithm of the concentration of hydronium ion in a solution i.e. $pH = \log \left[ {{H^ + }} \right]$”.
It is known to you that water self-ionizes itself into hydronium $({H^ + })$ and hydroxide $(O{H^ - }\ ions$: ${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$ and the equilibrium constant for this reaction is called ionization constant
$({K_w})$.
Also, it is known to use from experiment that ionic product of water ,${K_w}$ at 298K is $1 \times {10^{ - 14}}$ and is expressed as ${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$.
Since, water is neutral so it should have equal concentration of both ${H^ + }$ and $O{H^ - }$ ions.
Now, Calculation:
Let concentration of ${H^ + }$ be x.
${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]=1 \times {10^{ - 14}}$.
Since, from above it is pretty obvious that $\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]$,
\[
\Rightarrow x \times x = {10^{ - 14}}
\Rightarrow {x^2} = {10^{ - 14}}
\Rightarrow x = {10^{ - 7}}
\Rightarrow \left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]{10^{ - 7}}
\].
Therefore, from above calculations you can now clearly conclude that option A is the correct option to the given question.
Note: It should be remembered that the ionic product of water, ${K_w}$ increases with increase in temperature. This happens because of the fact that with increase, the degree of ionization of water increases i.e. more of molecules dissociate into ${H^ + }$ and $O{H^ - }$ ions. Thus, concentration of ${H^ + }$ and $O{H^ - }$ ions increases and hence, ionic products also increases.
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