Question

If Kw = 10-12 at 90°C, then the pH of pure water
and its nature at 90°C will be
(1) pH = 7, neutral . (2) pH = 6, acidic
(3) pH = 6, neutral (4) pH = 8, neutral​

Answer 1

Answer:

The pH of pure water is  6 neutral.

Explanation:

The equation shows the formation of hydroxonium and hydroxide ions from water:

$H_{2} O(l) \rightleftharpoons H^{+}(a q)+O H^{-}(a q)$

It is an endothermic reaction.  

The common method of determining any value is on the basis of the logarithmic scale of base 10. It is shown as:

$p_{x}=-\log x$

where X = Quantity of the desired and log = logarithm to base 10.

Thus, the expression for pH value is  

$p_{H}=-\log \left[H_{3} O^{+}\right]$

where $[H_3O^+]$=  the molar concentration of hydronium ion in the solution.

Taking antilog gives the value of hydronium ion molarity:

$10-p_{H}=\left[H_{3} O^{+}\right]$

Similarly, for pOH, it is expressed as:

$p O H=-\log \left[O H^{-}\right]$

Taking antilog on both sides:

$10-p O H=\left[O H^{-}\right]$

The relation of Kw with the two ion concentration is given by:

$K_{W}=\left[H_{3} O^{+}\right]\left[O H^{-}\right]$

Taking log and substituting the above values, we get

$-\log K_{w}=-\log \left(\left[H_{3} O^{+}\right]\left[O H^{-}\right]\right)=-\log \left[H_{3} O^{+}\right]+\left(-\log \left[O H^{-}\right]\right)$

$\mathrm{DK}_{\mathrm{W}}=\mathrm{pH}+\mathrm{pOH}$

This is the required equation.

Given data, at 90°C, $K_w$  = $1.0 \times 10^{-12},\text { and } \mathrm{so}:$

$K_{w}=\left[H_{3} O^{+}\right]\left[O H^{-}\right]$

Substituting the values, we get the pH and pOH of a neutral solution at this temperature as:

$\mathrm{K}_{\mathrm{W}}=1.0 \times 10^{-12}$

This implies,

$p H=-\log \left[H_{3} O^{+}\right]=-\log \left(1.0 \times 10^{-6}\right)=6.00$

$p O H=-\log [O H-]=-\log \left(1.0 \times 10^{-6}\right)=6.00$

So, the solution is neutral with pH at 6.