Question

Calculate pH of ${{10}^{-9}}M$ NaOH?

Answer 1

Hint: A scale used to specify the acidity or basicity of an aqueous solution in chemistry. The pH of acidic solutions is lower than that of basic or alkaline solutions. The pH scale is logarithmic, indicating the concentration of hydrogen ions in a solution in inverse order. Because the pH formula approximates the negative of the base 10 logarithm of the molar concentration of hydrogen ions in the solution, this is the case. pH is defined as the negative of the base 10 logarithm of the \[{{H}^{+}}\] ion's activity.

Complete answer:
At \[25\text{ }{}^\circ C\], acidic solutions have a pH less than 7, while basic solutions have a pH greater than 7. At this temperature, solutions with a pH of 7 are neutral. The pH neutral value changes with temperature, becoming lower than 7 as the temperature rises. For highly strong acids, the pH value can be less than 0 and for very strong bases, it can be more than 14. The pH scale may be traced back to a collection of standard solutions whose pH has been agreed upon internationally. The potential difference between a hydrogen electrode and a standard electrode, such as the silver chloride electrode, is measured in a concentration cell with transference to derive primary pH standard values.
The Henderson–Hasselbalch equation k can be used to estimate the pH of a buffer solution in chemistry and biology. The acid dissociation constant, \[{{K}_{a}}\] , has a numerical value that is known or
$pH=p{{K}_{a}}+{{\log }_{10}}\left( \dfrac{[Base]}{[Acid]} \right)$
assumed. For specified quantities of the acid, HA, and a salt, MA, of its conjugate base, \[{{A}^{}}\], the pH is determined; for example, the solution may include acetic acid and sodium acetate.

Consider the following two equilibria in solution:
$\mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{OH}^{-}$
and $\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$
From $\mathrm{NaOH}:\left[\mathrm{OH}^{-}\right]=10^{-9} \mathrm{M}$
From $\mathrm{H}_{2} \mathrm{O}:\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}$
Since $\left[\mathrm{OH}^{-}\right]_{\mathrm{H}_{2} \mathrm{O}}>>\left[\mathrm{OH}^{-}\right]_{\mathrm{NaOH}^{\prime}}$
The $\mathrm{pOH}$ will be determined from $\left[\mathrm{OH}^{-}\right]_{\mathrm{H}_{2} \mathrm{O}}=10^{-7} \mathrm{M}$
$\mathrm{pOH}=-\log \left(\left[\mathrm{OH}^{-}\right]\right)=-\log \left(10^{-7}\right)=7$
Therefore, to calculate the $\mathrm{pH}$ :
$\mathrm{pH}=\mathrm{pK}_{\mathrm{w}}-\mathrm{pOH}=14-7=7$
$\Rightarrow \text{pH}=7$
Hence 7 is the answer.

Note:
Many plants, including hibiscus, red cabbage (anthocyanin), and grapes, have pH-dependent plant pigments that can be utilised as pH monitors (red wine). Citrus juice is acidic mostly due to the presence of citric acid. Other carboxylic acids can be found in a wide range of living organisms. Lactic acid, for example, is generated by muscular action. Phosphate derivatives, such as ATP, have a pH-dependent condition of protonation. In a mechanism known as the Root effect, pH affects the activity of the oxygen-transport enzyme haemoglobin.