
The pH value of ordinary water is:
(A) 7
(B) 6.5
(C) 5.3
(D) 7.8
Answer
591k+ 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 at 25 degree Celsius. Now by knowing the ionization product of water we can easily solve this question.
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 ionization constant for 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 has equal concentration of both \[{H^ + }\] and \[O{H^ - }\] ions.
Now, Calculation of pH of water:
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 \[\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]\],
\[
\Rightarrow x \times x = {10^{ - 14}} \\
\Rightarrow {x^2} = {10^{ - 14}} \\
\Rightarrow x = {10^{ - 7}} \\
\left[ {{H^ + }} \right] = {10^{ - 7}} \\
\]
\[
pH = \log \left[ {{H^ + }} \right] \\
\Rightarrow pH = - \log {10^{ - 7}} \\
\Rightarrow pH = - ( - 7)\log 10 \\
pH = 7 \\
\]
Therefore, above calculations we can now clearly conclude that option A is the correct option to the given question.
Note: It should be remembered to you that those substances which have pH less than 7 at 298K are known as acids and those who have pH more than 7 are termed as base.
Also, it should be remembered that pH measurement has a wide variety of applications, one of which is in agriculture. By measuring the pH of soil, you can easily decide which crop will grow readily in that soil
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 ionization constant for 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 has equal concentration of both \[{H^ + }\] and \[O{H^ - }\] ions.
Now, Calculation of pH of water:
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 \[\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]\],
\[
\Rightarrow x \times x = {10^{ - 14}} \\
\Rightarrow {x^2} = {10^{ - 14}} \\
\Rightarrow x = {10^{ - 7}} \\
\left[ {{H^ + }} \right] = {10^{ - 7}} \\
\]
\[
pH = \log \left[ {{H^ + }} \right] \\
\Rightarrow pH = - \log {10^{ - 7}} \\
\Rightarrow pH = - ( - 7)\log 10 \\
pH = 7 \\
\]
Therefore, above calculations we can now clearly conclude that option A is the correct option to the given question.
Note: It should be remembered to you that those substances which have pH less than 7 at 298K are known as acids and those who have pH more than 7 are termed as base.
Also, it should be remembered that pH measurement has a wide variety of applications, one of which is in agriculture. By measuring the pH of soil, you can easily decide which crop will grow readily in that soil
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