Question

The pH value of pure water at $300K$ is
A. Exactly $7$
B. Slightly >7
C. Zero
D. Slightly <7

Answer 1

We have to remember that the pH is the measure of hydrogen concentration in a solution. It can be calculated by the formula $pH = - \log \left( {{H^ + }} \right)$ (or) $p{K_w} = pH + pOH = 14$ .It scales from 0 to 14. If the pH of the solution ranges from 0 to 7 then the solution is said to be acidic, if the pH of the solution ranges from above 7 to 14 then the solution is said to be basic, if the pH of the solution is equal to $7$ then it is neutral.

Complete step by step answer:
Now let us consider the dissociation of water is,
${H_2}O \rightleftarrows O{H^ - } + {H^ + }$
In the above equation the concentration of ${H^ + }$ determines the acidic strength of water. When there is a change in temperature of the solution, dissociation of the solution also changes. Here, as the temperature of the solution increases, the dissociation of ${H_2}O$ will lead to the increase of ${H^ + }$ ions in water. So, there will be change in pH. As the temperature increases the concentration of ${H^ + }$ ion decreases i.e., slight lower in pH.

Therefore, the option D is correct.

Note: We have to remember that according to Le chatlier principle, if you increase the temperature of water there will lower in dissociation water, it can be given as pH in inversely proportional to temperature.
$pH$ can be calculated by the formula. At $296K$ or $25^\circ C$ the pH of water is
$pH = \dfrac{1}{2}p{K_w}$
At $25^\circ C$ the ${K_w}$ of water is $1.00 \times {10^{ - 14}}mo{l^2}d{m^{ - 6}}$
Now let us calculate the value of $p{K_w}$
$p{K_w} = - {\operatorname{logK} _w}$
Now substitute the value $K_w$
$p{K_w} = - \log ({\text{1}}{\text{.00}} \times {\text{1}}{{\text{0}}^{ - 14}})$
Take log of ${K_w}$
$p{K_w} = - ( - 14)$
$p{K_w} = 14$
The relationship between $p{K_w}$ and pH is given by the equation
$pH = \dfrac{1}{2}p{K_W}$
Now substitute the value of $p{K_w}$ in the above equation
$pH = \dfrac{1}{2}(14)$
On simplification we get,
$pH = 7$
At $300K$ or $30^\circ C$ the pH of water is
$pH = \dfrac{1}{2}p{K_w}$
At $300K$ or $30^\circ C$ the ${K_w}$ of water is $6.92 \times {10^{ - 14}}{\text{ mo}}{{\text{l}}^2}d{m^{ - 6}}$
Now let us calculate the value of $p{K_w}$,
$p{K_w} = - {\operatorname{logK} _w}$
Now substitute the value KW
$p{K_w} = - \log (6.92 \times {\text{1}}{{\text{0}}^{ - 14}})$
Take log of ${K_w}$,
$p{K_w} = - ( - 13.83)$
$p{K_w} = 13.83$
The relationship between $p{K_w}$ and pH is given by the equation
$pH = \dfrac{1}{2}p{K_W}$
Now substitute the value of $p{K_w}$ in the above equation
$pH = \dfrac{1}{2}(13.83)$
On simplification we get,
$pH = 6.92$
From the above two example, we see that temperature increases pH value decreases that is at ${25^0}C$, the $pH = 7$
At ${30^0}{\text{C}}$ , The $pH = 6.92$ , the temperature at $300K$ or $30^\circ C$ is a slight increase from ${25^0}C$ . So we conclude that there will be a slight decrease in pH from $7$.