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Question: What is pOH of an aqueous solution with \[\left[ {{H}^{+}} \right]={{10}^{-2}}\text{ }M,\text{ }Kw=2...

What is pOH of an aqueous solution with [H+]=10−2 M, Kw=2×10−12\left[ {{H}^{+}} \right]={{10}^{-2}}\text{ }M,\text{ }Kw=2\times {{10}^{-12}} ? Report your answer after dividing by two and round it off to the nearest whole number. Percentage ionization of water as follows at a certain temperature is 3.6×10−73.6\times {{10}^{-7}}%. Calculate Kw and pH of water temperature.

Explanation

Solution

We know that the Kw is known as the coefficient of the water which can be determined by using the concentration of H+{{H}^{+}} and OH−O{{H}^{-}} ions which are present in the given solution. A low ionization degree is at times referred to as partially or weakly ionized, and a high degree of ionization as fully ionized.

Complete answer:
Few bases such as sodium hydroxide or lithium hydroxide get completely dissociated into their ions in an aqueous solution, which is referred to as the strong bases. The base ionization constant, that is, Kw, is referred to as an equilibrium constant for the ionization of a base. Thus, we can say that a strong base implies a good proton acceptor, whereas a strong acid implies a good proton donor. The dissociation of either weak acids or weak bases in water. As we know that the [H+][OH−]=Kw.\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{K}_{w}}.
Here by rearranging the [OH−]=Kw[H+]\left[ O{{H}^{-}} \right]=\dfrac{Kw}{\left[ {{H}^{+}} \right]} on substitution we get; [OH−]=2×10−1210−2=2×10−10\left[ O{{H}^{-}} \right]=\dfrac{2\times {{10}^{-12}}}{{{10}^{-2}}}=2\times {{10}^{-10}}
Now, pOH=−log⁡[OH−]−log⁡102×10−10pOH=-\log \left[ O{{H}^{-}} \right]-{{\log }_{10}}2\times {{10}^{-10}} on further solving we get;
⇒−log⁡1010−10+log⁡102\Rightarrow -\\{{{\log }_{10}}{{10}^{-10}}+\log _{10}^{2}\\}
By substituting the values, we get;
⇒−−10+0.301\Rightarrow -\\{-10+0.301\\}
⇒10−0.301⇒9.699\Rightarrow 10-0.301\Rightarrow 9.699
By taking half of the pOH, we get the required value
pOH2=9.6992=4.8495≈5.\dfrac{pOH}{2}=\dfrac{9.699}{2}=4.8495\approx 5.
Although, the ionization degree is also called an ionization yield, which refers to the proportion of neutral particles, like those in aqueous or gaseous solutions, which are ionized to the charged particles. For electrolytes, it could be understood as the capacity of either acid or base to ionize itself.

Note:
Remember that the ionization degree of acids and bases depends on the degree of dissociation of compounds into their constituent ions. The strong acids and bases have a high degree of ionization when compared to the ionization of weak acid and base. Also, a strong acid implies a good proton donor, whereas a strong base implies a good proton acceptor.