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Question: The value of \[{K_w}\] at the physiological temperature (\[{37^o}C\]) is \(2.56 \times {10^{ - 14}}\...

The value of Kw{K_w} at the physiological temperature (37oC{37^o}C) is 2.56×10−142.56 \times {10^{ - 14}}. What is the pH at the neutral point of water at this temperature? (log2=0.3log2 = 0.3)

Explanation

Solution

To calculate the pH of any solution, this formula is used i.e. −log [H+] - log{\text{ }}\left[ {{H^ + }} \right]. So the very first thing we need is the concentration of H+{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⇌cC+dD  aA + bB \rightleftharpoons cC + dD\;
The equilibrium constant can be expressed as follows:
K=[C]c[D]d[A]a[B]bK = \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. H2O+H2O→OH−+H3O+{H_2}O + {H_2}O \to O{H^ - } + {H_3}{O^ + }, equilibrium constant becomes:
K=[OH−][H3O+][H2O]2 ∴K.[H2O]2=[OH−][H3O+]  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.Kw{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:
Kw=[OH−][H3O+]=[OH−][H+]{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. [OH−]=[H+][O{H^ - }] = [{H^ + }]
Kw{K_w}= 2.56×10−142.56 \times {10^{ - 14}}at 37oC{37^o}C (Given)
We know that Kw=[OH−][H+]{K_w} = [O{H^ - }][{H^ + }]
As [OH−]=[H+][O{H^ - }] = [{H^ + }], thus:
Kw=[H+][H+]=[H+]2 [H+]=Kw12=(2.56×10−14)12 [H+]=1.6×10−7  {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 [H+] - log{\text{ }}\left[ {{H^ + }} \right]
pH=−log [1.6×10−7]=6.795=6.8pH = - log{\text{ }}\left[ {1.6 \times {{10}^{ - 7}}} \right] = 6.795 = 6.8

Hence, the pH at the neutral point of water at 37oC{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.