Question

The Ionic product of water at $310K$ is $2.7 \times {10^{ - 14}}$.what is the $pH$of neutral water at this temperature?

Answer 1

When we talk about the ionic product that means the product should have positive as well as negative ions present in the product. $pH$ stands for the potentia of hydrogen. The formulas we will use here are represented as:
${K_w} = \dfrac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}}$ and $pH = \log \left[ {{H^ + }} \right]$

Complete step-by-step answer:
We are given the ionic product of water to be $2.7 \times {10^{ - 14}}$.
Also, the temperature given is $310K$
Now as we have seen in the question that an ionic product is being talked about now we will discuss what is an ionic product of water here. When we talk about the ionic product that means the product should have positive as well as negative ions present in the product. Basically the ionic product of water contains both positive hydronium ions and negatively charge on hydroxide ions and is shown by the formula:
${K_w} = \dfrac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{H_2}O} \right]}}$ such that ${K_w}$is the ionic product of water.
Now, let us further talk about the pH of neutral water that is basically the power of hydrogen. We need to find the pH of the neutral water at the given temperature.
In the question, we are given ${K_w}$as $2.7 \times {10^{ - 14}}$
Also as we are talking about the neutral water the hydrogen ions should be equal to hydronium ions.
Then we should know that $pH = \log \left[ {{H^ + }} \right]$
So, we firstly need to calculate the $\left[ {{H^ + }} \right]$ ions value.
Then from the above statement about the neutrality we come to know that the $\left[ {{H^ + }} \right]$ is half the value of ${K_w}$ .
So, the concentration of hydrogen ions are: \left[ {{H^ + }} \right]$$$$ = \sqrt {2.7 \times {{10}^{ - 14}}} = 1.7 \times {10^{ - 7}}
As we know that$pH = \log \left[ {{H^ + }} \right]$
putting the value of $\left[ {{H^ + }} \right]$ in it we get
$pH = \log \left( {1.7 \times {{10}^{ - 7}}} \right) = 6.7$
The above expression is the required answer.

Note: $pH$ is always measured using $pH$ paper or the $pH$ meter. $pH$ has a wide application in the agriculture department as it is used to measure the acidity or the basicity of the soil and also used in the water treatment plants to check the quality of water.