
How do you find $pOH$ given $pH$?
Answer
560.4k+ views
Hint:As we all know that $pH$ of a solution is basically the negative logarithm of hydrogen ion concentration and $pOH$ is calculated by knowing the hydroxide ion concentration in the solution and is given as the negative logarithm of hydroxide ions concentration.
Complete step by step solution:
As we have already discussed that $pH$ and $pOH$ both are the negative logarithms of the concentrations of hydrogen ions and hydroxide ions respectively.
We know that, $pH$ is expressed in terms of hydrogen ion concentration as shown below:
$pH = - \log [{H^ + }]$
We also know that, $pOH$ is expressed in terms of hydroxide ion concentration as shown below:
$pOH = - \log [O{H^ - }]$
Now, we know that water generally dissociates into hydrogen ions and hydroxide ions as shown:
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
Then the equilibrium constant is given as ratio of the concentration of products to the concentration of the reactant. So it will be:
$K = \dfrac{{[{H^ + }][O{H^ - }]}}{{[{H_2}O]}}$
Now, in terms of ionic product of water we can write the above equation as:
$K[{H_2}O] = {K_W}$
In other words, we can say that the ionic product of water is equivalent to the product of concentrations of hydrogen ion and hydroxide ion and it will be written as:
${K_W} = [{H^ + }][O{H^ - }]$
We are also aware with the fact that the hydrogen ion and hydroxide ion concentration are related as $14$ in the solution.
So we can write:
$[{H^ + }][O{H^ - }] = 14$
In terms of $pH$ and $pOH$ we can write the above equation as:
$pH + pOH = 14$
Or
$pOH = 14 - pH$
Therefore, we can find the value of $pOH$ by simply subtracting the given value of $pH$ from $14$.
Note:Remember that when the hydrogen ion concentration and hydroxide ion concentrations are equal then the solution is generally neutral and when the hydroxide ion concentration is more in the solution, then the solution is basic in nature and when the hydrogen ion concentration is more, then the solution is acidic in nature which can be identified with the help of the formula below: $pH + pOH = 14$.
Complete step by step solution:
As we have already discussed that $pH$ and $pOH$ both are the negative logarithms of the concentrations of hydrogen ions and hydroxide ions respectively.
We know that, $pH$ is expressed in terms of hydrogen ion concentration as shown below:
$pH = - \log [{H^ + }]$
We also know that, $pOH$ is expressed in terms of hydroxide ion concentration as shown below:
$pOH = - \log [O{H^ - }]$
Now, we know that water generally dissociates into hydrogen ions and hydroxide ions as shown:
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
Then the equilibrium constant is given as ratio of the concentration of products to the concentration of the reactant. So it will be:
$K = \dfrac{{[{H^ + }][O{H^ - }]}}{{[{H_2}O]}}$
Now, in terms of ionic product of water we can write the above equation as:
$K[{H_2}O] = {K_W}$
In other words, we can say that the ionic product of water is equivalent to the product of concentrations of hydrogen ion and hydroxide ion and it will be written as:
${K_W} = [{H^ + }][O{H^ - }]$
We are also aware with the fact that the hydrogen ion and hydroxide ion concentration are related as $14$ in the solution.
So we can write:
$[{H^ + }][O{H^ - }] = 14$
In terms of $pH$ and $pOH$ we can write the above equation as:
$pH + pOH = 14$
Or
$pOH = 14 - pH$
Therefore, we can find the value of $pOH$ by simply subtracting the given value of $pH$ from $14$.
Note:Remember that when the hydrogen ion concentration and hydroxide ion concentrations are equal then the solution is generally neutral and when the hydroxide ion concentration is more in the solution, then the solution is basic in nature and when the hydrogen ion concentration is more, then the solution is acidic in nature which can be identified with the help of the formula below: $pH + pOH = 14$.
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