Question

Given that ${K_W}$ for water is ${10^{ - 13}}{M^2}$ at${2^ \circ }C$, compute the sum of $pOH$ and $pH$ for a neutral aqueous solution at ${2^ \circ }C$?
A. 7.0
B. 13.30
C. 14.0
D. 13.0

Answer 1

Hint: ${K_w} = [{H^ + }][O{H^ - }]$
In neutral solution, $[{H^ + }] = [O{H^ - }]$
Formula used:
$p{K_w} = pH + pOH$

Complete step by step answer:
Ionic products of water (${K_W}$) may be defined as the product of molar concentrations of ${H^ + }$ ions and $O{H^ - }$ ions.
Therefore, ${K_w} = [{H^ + }][O{H^ - }]$
As ${H^ + }$ ions in water exists as ${H_3}{O^ + }$ ions , it can be written as
${K_w} = [{H_3}{O^ + }][O{H^ - }]$
Therefore, ionic products of water can also be defined as the product of molar concentration of ${H_3}{O^ + }$ ions and $O{H^ - }$ ions.
$pH$ may be defined as a negative logarithm of hydronium ion concentration.
$pH = - \log [{H_3}{O^ + }] = - \log [{H^ + }]$
Similarly, we have
$pOH = - \log [O{H^ - }]$
$p{K_w} = - \log {K_w}$
For a neutral solution, $pH = pOH$
Now, $p{K_w} = pH + pOH$
It is given that${K_w} = {10^{ - 13}}{M^2}$at ${2^ \circ }C$
$
   \Rightarrow pH + pOH = p{K_w} = - \log ({10^{ - 13}}) \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - ( - 13)({\log _{10}}10) \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 13 \\
  $
Therefore, the sum of $pH$ and $pOH$ for a neutral aqueous solution at ${2^0}C$ is $13.0$.

Hence, the option (D) is the correct answer.

Note:
Ionic products of water are constant only at constant temperature.
The value of ${K_W}$ is usually taken as $1.008 \times {10^{ - 14}}mo{l^2}{L^{ - 2}}$at $298K$.
The $[{H_3}{O^ + }]$ and $[{H^ + }]$ions are always present in an aqueous solution. But their relative concentrations are different in different types of solutions.
So, in a neutral solution, $[{H_3}{O^ + }] = [O{H^ - }]$
In acidic solution, $[{H_3}{O^ + }] > [O{H^ - }]$
In alkaline solution, $[{H_3}{O^ + }] < [O{H^ - }]$