Question

The value of \[{K_w}\] at the physiological temperature (\[{37^o}C\]) is $2.56 \times {10^{ - 14}}$. What is the pH at the neutral point of water at this temperature? (\[log2 = 0.3\])

Answer 1

Hint: To calculate the pH of any solution, this formula is used i.e. \[ - log{\text{ }}\left[ {{H^ + }} \right]\]. So the very first thing we need is the concentration of \[{H^ + }\] ions in the solution.

Complete step by step answer:
The equilibrium constant, denoted by K, expresses the relationship between reactants and products of a reaction at an equilibrium condition with respect to a specific unit. For a generalised chemical reaction taking place in a solution:
\[aA + bB \rightleftharpoons cC + dD\;\]
The equilibrium constant can be expressed as follows:
$K = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
where [A], [B], [C] and [D] refer to the molar concentration of species A, B, C, D respectively at equilibrium. The coefficients like a, b, c, and d in the generalised chemical equation become exponents as seen in the above expression.
In case of auto-protolysis reaction of water i.e. ${H_2}O + {H_2}O \to O{H^ - } + {H_3}{O^ + }$, equilibrium constant becomes:
$
  K = \dfrac{{[O{H^ - }][{H_3}{O^ + }]}}{{{{[{H_2}O]}^2}}} \\
  \therefore K.{[{H_2}O]^2} = [O{H^ - }][{H_3}{O^ + }] \\
$
This is known as the ionization constant of water i.e.\[{K_w}\]which actually either refers to the ionic product of water or the product of hydroxide and hydrogen ions that are present in a solution which means:
 ${K_w} = [O{H^ - }][{H_3}{O^ + }] = [O{H^ - }][{H^ + }]$
This states that the ionic product of water equals the product of hydroxide and hydrogen ions that are present in a solution.
In the question it is given that water is at neutral point according to which concentration of hydroxide and hydrogen ions should be equal i.e. $[O{H^ - }] = [{H^ + }]$
\[{K_w}\]= $2.56 \times {10^{ - 14}}$at \[{37^o}C\] (Given)
We know that ${K_w} = [O{H^ - }][{H^ + }]$
As $[O{H^ - }] = [{H^ + }]$, thus:
$
  {K_w} = [{H^ + }][{H^ + }] = {[{H^ + }]^2} \\
  [{H^ + }] = {K_w}^{\dfrac{1}{2}} = {(2.56 \times {10^{ - 14}})^{\dfrac{1}{2}}} \\
  [{H^ + }] = 1.6 \times {10^{ - 7}} \\
$
Using this value of hydrogen ion concentration, we can find the value of pH using the formula:
pH =\[ - log{\text{ }}\left[ {{H^ + }} \right]\]
$pH = - log{\text{ }}\left[ {1.6 \times {{10}^{ - 7}}} \right] = 6.795 = 6.8$

Hence, the pH at the neutral point of water at \[{37^o}C\] is 6.8.

Note:The pH scale ranges from 0 to 14, and most solutions fall within this range. Any solution lying below 7.0 is considered to be acidic while above 7.0 is alkaline. If the solution has a pH of 7 then it is considered to be neutral.