Question

At ${100^0}C$ the ${K_M}$ of the water is 55 times its value at ${25^0}C$.What will be the pH of the neutral solution? ($\log (55) = 1.74$)
A.6.13
B.7.00
C.7.87
D.5.13

Answer 1

Hint: To answer this question, we first need to understand what IS pH. In a water-based solution, pH is the negative log of hydrogen ion concentration. Sren Peter Lauritz Srensen, a Danish biologist, was the first to coin the term "pH" in 1909. pH stands for "potential of hydrogen," with "p" standing for the German word for power, potenz, and H standing for the element symbol for hydrogen.

Complete answer:
As we know that at ${25^0}C,{K_M} = 1 \times {10^{ - 14}}$
And as given in the question that at ${100^0}C,{K_M} = 55 \times {10^{ - 14}}$
So, by using the formula that
$[{H^ + }] = \sqrt {55 \times {{10}^{ - 14}}} $
And formula of pH is,
$pH = - \log [{H^ + }]$
And putting the value of $[{H^ + }]$ concentration
$pH = - \log [\sqrt {55 \times {{10}^{ - 14}}} ]$
We get
$\rightarrow pH = \dfrac{1}{2}[ - \log (55) + 14\log (10)]$
$\rightarrow pH = \dfrac{1}{2}[ - 1.74 + 14]$
$pH = 6.13$
So, the final answer is option(A) i.e., 6.13.

Note:
The pH meter is frequently referred to as a "potentiometric pH meter" because it monitors the difference in electrical potential between a pH electrode and a reference electrode. The acidity or pH of the solution is related to the difference in electrical potential.