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Question: One liter of water contains \[{{10}^{-7}}\]moles of \[{{H}^{+}}\]ions. Degree of ionization of water...

One liter of water contains 107{{10}^{-7}}moles of H+{{H}^{+}}ions. Degree of ionization of water is:
A. 1.8×1071.8\times {{10}^{-7}}
B. 0.8×1090.8\times {{10}^{-9}}
C. 5.4×1095.4\times {{10}^{-9}}
D. 5.4×1075.4\times {{10}^{-7}}

Explanation

Solution

Degree of ionization is the ratio of ionized fraction of the solute to the total amount of dissolved solute and it is going to be denoted with a symbolα\alpha .
 !!α!! =number of moles ionizedTotal number of moles of solute\text{ }\\!\\!\alpha\\!\\!\text{ =}\frac{\text{number of moles ionized}}{\text{Total number of moles of solute}}
Complete step by step answer:
In the question it is given that one liter water contains 107{{10}^{-7}}moles of H+{{H}^{+}} ions.

& \underset{concentration\text{ }At\text{ }initial}{\mathop{~}}\,~\text{ }\underset{C}{\mathop{{{H}_{2}}O}}\,\to \underset{0}{\mathop{{{H}^{+}}}}\,+\underset{0}{\mathop{O{{H}^{-}}}}\, \\\ & ~\underset{After\text{ }dissociation}{\mathop{~~}}\,~\text{ }\underset{C-C\alpha }{\mathop{{{H}_{2}}O}}\,\to \underset{C\alpha }{\mathop{{{H}^{+}}}}\,+\underset{C\alpha }{\mathop{O{{H}^{-}}}}\,~~~\text{ } \\\ & ~~~~~\text{ } \\\ \end{aligned}$$ We know that the formula of Ionic product of water $${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$$ $$[{{H}^{+}}]$$= Concentration of $${{H}^{+}}$$ion $$[O{{H}^{-}}]$$= concentration of $$O{{H}^{-}}$$ion $$[{{H}^{+}}]=\frac{{{10}^{-7}}moles}{1L}={{10}^{-7}}M$$ Therefore, C$$\alpha $$= $${{10}^{-7}}M$$ We know that density of pure water = 1g/ml Number of moles of water = $$\frac{\text{given mass}}{\text{molar mass}}$$ Here given mass = 1g Molar mass of water = 18 g Number of moles of water = $$\frac{\text{given mass}}{\text{molar mass}}=\frac{1}{18}=0.055moles$$ We know that volume of water 1 ml = 0.001 L Therefore the concentration of water C = $$[{{H}_{2}}O]=\frac{moles}{\text{Volume in L}}=\frac{0.0555}{0.001L}=55.5M$$ Substitute C in C$$\alpha $$. $\begin{aligned} & 55.5 \ \alpha = 10^{-7} \ M \\\ & \alpha = \frac{{10^{-7}}}{55.5}= 0.018 \times 10^{-7} \\\ & \alpha\% = 1.8 \times 10^{-7}\% \\\ \end{aligned}$ **So, the correct option is A.** **Additional information:** An acid or a base strength denotes their degree of ionization. A strong acid will completely ionize or dissociate in water while a weak acid ionizes partially. A strong base will completely ionize or dissociate in water while a weak base ionizes partially. **Note:** Don’t be confused with degree of ionization or degree of dissociation. Both are the same. Ionization means the ions present in the molecule are going to convert into positive and negative ions in a particular solvent. This type of process is called dissociation or ionization of the molecule.