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-Firstly, we have to calculate the concentration of water i.e no. of a mole per litre. Here, we know that the number of moles of water is 1000/18 as the molar mass of the water is 18 $(1\text{ }\times \text{ 2) + 16 = 18}$ and the mass of water is 1000g.

-So, water has the concentration = $\dfrac{1000}{18}=\text{ 55}\text{.66 mol }{{\text{L}}^{-1}}$.

-Now, we that a small amount is dissociated and give hydrogen ion and hydroxyl ion i.e.

${{\text{H}}_{2}}\text{O }\to \text{ }{{\text{H}}^{+}}\text{ + O}{{\text{H}}^{-}}$

-Here, hydrogen ion and hydroxyl ion are yielded in equal concentration. So, the ionic product of the water (\[{{\text{K}}_{\text{W}}}\]) will be:

\[{{\text{K}}_{\text{W}}}\text{ = }\left( {{\text{H}}^{+}} \right)\left( \text{O}{{\text{H}}^{-}} \right)\text{ = }{{\text{C}}^{2}}{{\alpha }^{2}}\]

-Here, C is the concentration of water and alpha is the degree of dissociation of water which is given in the question i.e. $1.8\text{ }\times \text{ 1}{{\text{0}}^{-9}}$

$\begin{align}

& {{\text{K}}_{\text{W}}}\text{ = (55}\text{.56}{{\text{)}}^{2}}\text{ }\times \text{ (1}\text{.8 }\times \text{ 1}{{\text{0}}^{-9}}{{)}^{2}} \\

& \text{ = 1 }\times \text{ 1}{{\text{0}}^{-14}} \\

\end{align}$

-So, the ionic product of the water is $\text{1}{{\text{0}}^{-14}}$.

-Now, we have to calculate the ionic constant of the water, which is represented by \[{{\text{K}}_{\text{a}}}\], it will be

\[{{\text{K}}_{\text{a}}}\text{ = }\dfrac{({{\text{H}}^{+}})(\text{O}{{\text{H}}^{-}})}{({{\text{H}}_{2}}\text{O)}}\]

=\[\dfrac{{{\text{C}}^{2}}{{\alpha }^{2}}}{\text{C}\alpha }\] here, the value of alpha of water is so much smaller than 1, so it can be neglected. The equation will now become:

\[\text{C}{{\alpha }^{2}}\text{ = 55}\text{.56 }\times \text{ (1}\text{.8 }\times \text{ 1}{{\text{0}}^{-9}}{{)}^{2}}\] \[\text{= 1}\text{.8 }\times \text{ 1}{{\text{0}}^{-16}}\].

Therefore, the value of the ionic constant and ionic product of water is $\text{1}{{\text{0}}^{-14}}$ and \[\text{1}\text{.8 }\times \text{ 1}{{\text{0}}^{-16}}\] respectively.