Question

In pure water, some of the molecules ionize according to the equation H2O→H+ + OH−. The extent of the ionization increases with temperature. A student heats pure water and records the measured pH at 50°C as 6.6. Based on this information, what mathematical relationships gives the pOH pOH of pure water at 50°C?

Answer 1

Answer : The mathematical relationships gives the pOH of pure water at $50^oC$ is, $pOH=14-pH$

Explanation :

As we are given that the pH at $50^oC$ is, 6.6.

Now we have to determine the mathematical relationships gives the pOH of pure water at $50^oC$.

As we know that:

The ionization of water is:

$H_2O\overset{K_w}\rightarrow H^++OH^-$

The expression for dissociation constant for water is:

$K_w=[H^+][OH^-]$

taking logarithm on both side, we get:

$\log K_w=\log [H^+]+\log [OH^-]$

Taking negative sign on both side, we get:

$-\log K_w=-\log [H^+]+(-\log [OH^-])$

$pK_w=pH+pOH$

As we know that the value of $pK_w$ is 14 at 25-50°C.

So,

$14=pH+pOH$

or,

$pOH=14-pH$

$14=6.6+pOH$

$pOH=7.4$

Therefore, the mathematical relationships gives the pOH of pure water at $50^oC$ is, $pOH=14-pH$