Question

The value of $Kw$​ is $9.55 \times {10^{ - 14}}$ at a certain temperature. Calculate the pH of water at this temperature.

Answer 1

Hint: $Kw$ is called the coefficient of water that can be calculated using the concentration of ${H^ + }$ and $O{H^ - }$ present in the solution. For water, ${H^ + }$= $O{H^ - }$ . Using this relation we can find the answer to this question.

Formula used :Kw is the equilibrium constant.
$Kw = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$
$pH = - \log \left[ {{H^ + }} \right]$
Kw is the equilibrium constant. Where $\left[ {{H^ + }} \right]$ and $\left[ {O{H^ - }} \right]$ is the concentration of hydrogen ions and hydroxide ions respectively.

Complete step by step answer:
The above coefficient $Kw$ is called the autoionization constant of water. It has a magnitude of the order of ${10^{ - 14}}$ which is derived from the pH and pOH of the solution. For
For this question we know the value of $Kw$ and by plugging it in the formula mentioned above we get the following set of equation:
$Kw = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$
Since, we know the relation given below we get,
$\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]$
$Kw = \left[ {{H^ + }} \right]\left[ {{H^ + }} \right]$
$Kw = {\left[ {{H^ + }} \right]^2}$
substituting the value of ionization constant of water, we get,
$9.55 \times {10^{ - 14}} = {\left[ {{H^ + }} \right]^2}$
$\sqrt {9.55 \times {{10}^{ - 14}}} = \left[ {{H^ + }} \right]$
therefore, the concentration of hydrogen ion is,
$3.09 \times {10^{ - 7}} = \left[ {{H^ + }} \right]$

Now that we have found the concentration of the ${H^ + }$ in the solution. we can find the pH.

$pH = - \log \left[ {{H^ + }} \right]$
$ = - \log \left[ {3.09 \times {{10}^{ - 7}}} \right]$
solving the logarithm we get,
$ = - \left[ {\log \left( {3.09} \right) + \log \left( {{{10}^{ - 7}}} \right)} \right]$
$ = - \left[ {0.49 - 7} \right]$
$ = - \left[ { - 6.55} \right]$
$pH = 6.55$

Thus, we have found the pH of the solution at that particular temperature that is,$6.55$ . This shows that water is slightly acidic.


Note: $Kw$ :It is temperature dependent. Meaning its value changes with temperature. P $Kw$ can also be found using the pH and pOH using the formula below:
$pKw = pH + pOH$

Since the pH and pOH of any aqueous solution at $298K$ is $7$ if the molar concentration of both ions is the same. Therefore, the $pKw$ is $14$ for any solution at $298K$ .

It is also important to remember that the pH can change with temperature.

The contribution of p $Kw$ is extremely significant in cases when the solution is extremely dilute and can influence the ionization of the solution.

$Kw$ is also known as the ionic product of water. $Kw$ increases with increase in temperature.
Remember that for pure water, we have to consider the ${H^ + }$ and $O{H^ - }$ to be equal.