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

At 25°C, the pH and pOH of pure water are both 7.00, but this changes at different temperatures.

Step-by-step explanation:

In pure water at 25 degrees Celsius, the pOH is equal to 7. This is because the pOH is the negative logarithm (base 10) of the hydroxide ion concentration in moles per liter, and in neutral water, the concentration of hydroxide ions (OH-) is equal to
`1.0 × 10^(-7) M.`

The pOH of pure water at 50°C can be determined using the relationship between pH and pOH. In this case, the hydronium ion molarity in pure water is
`1.0 x 10-7M`, which corresponds to a pH and pOH of 7.00. However, since the autoionization constant Kw is temperature dependent, the relationship changes at different temperatures. For example, at 80°C, the pH and pOH of pure water are 6.31.

Answer 2

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`