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In the question I've to find out the ionic product of water using the facts that the degree of dissociation of water at $18\ \mathrm{^\circ C}$ is $1.8\times10^{-9}$.

My attempt

Let the concentration of water be $1\ \mathrm M$.

$$\ce{H2O <=> H+ + OH-}$$

Initial:

$$\begin{align} [\ce{H2O}] &= 1\ \mathrm M\\[6pt] [\ce{H+}] &= 0\ \mathrm M\\[6pt] [\ce{OH-}] &= 0\ \mathrm M \end{align}$$

At equilibrium:

$$\begin{align} [\ce{H2O}] &= (1 - 1.8×10^{-9})\ \mathrm M\\[6pt] [\ce{H+}] &= 1.8×10^{-9}\ \mathrm M\\[6pt] [\ce{OH-}] &= 1.8×10^{-9}\ \mathrm M \end{align}$$

Therefore, $$\begin{align} K_\mathrm w &= \frac{[\ce{H+}] \times [\ce{OH-}]}{[\ce{H2O}]}\\[6pt] &= \left(1.8\times10^{-9}\right) \times \left(1.8\times10^{-9}\right)\\[6pt] &= 3.24 \times 10^{-18} \end{align}$$

According to my book answer is $1.0\times10^{-14}$, which I know is correct. I want to know where am I going wrong?

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    $\begingroup$ why concentration of water be assumed 1M? $\endgroup$
    – JM97
    Commented Aug 26, 2016 at 15:22
  • $\begingroup$ @JM97 They've asked to find the ionic product, so I thought that assuming the initial concentration of water to be 1M will simplify the problem. But I've done something wrong I guess. I don't know where am I going wrong. $\endgroup$ Commented Aug 26, 2016 at 15:32
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    $\begingroup$ molarity is number of moles per litre so please consider the number of moles of water present per liter . Hint: 1kg of water occupies one litre space. $\endgroup$
    – JM97
    Commented Aug 26, 2016 at 15:34
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    $\begingroup$ @JM97 There is no such thing as the "number of moles"; it is called amount of substance, just like mass is not called "number of kilograms". $\endgroup$ Commented Apr 5, 2018 at 11:34

2 Answers 2

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What it boils down to is, you need to take the fractional dissociation $1.8×10^{-9}$ and multiply it by the total amount of water, not just one molar. The total concentration of water is $1000$ grams per liter, which you convert to moles per liter by dividing by the molecular weight of the water. After working this out and doing the multiplication you should get the right hydron and hydroxide ion concentrations and then you get the right $K_w$.

This is just a number crunching exercise. To find the ion product for real, we can use electrochemistry. Using a highly reversible cathodic reaction for hydrogen evolution, say on a platinum electrode, find the hydrogen reduction potential for 1 M strong acid and for 1 M strong base, e.g. hydrochloric acud and sodium hydroxide. There is a difference between them, which you plug into the Nernst Equation to get the hydron concentration in the base solution. The ion product then follows.

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  • $\begingroup$ The concentration of water in pure water is $55.5$ M. So that the concentration of $\ce{H+}$ is $\pu{55.5 \times 1.8\times 10^{-9} = 1.00 \times 10^{-7}}$ M. $\endgroup$
    – Maurice
    Commented Jan 28 at 15:04
  • $\begingroup$ The problem is that percent dissociation does not give a molar concentration hence Maurice's treatment to convert to moles/L. % dissociation is grams/gram. Since one mole of water gives one mole of H+ 55.5 moles/L gives 55.5 times as many moles H+/L. [Formation of H3O+ does not affect the stoichiometry only one water disassociates.] Working out the mass balance between H3O+ and OH- is interesting. Exact molecular weights must be used and the mass of the electron taken into account. $\endgroup$
    – jimchmst
    Commented Jan 28 at 23:36
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What you did wrong was adding the concentration of the hydron and the hydroxide ion. Also, although it hardly matters, the concentration of $\ce{H_2O}$ is considered a constant and ignored in the product.

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  • $\begingroup$ That seems to be a typo which I fixed. It should have been apparent on reading the line following it, where they have indeed multiplied the concentrations. $\endgroup$
    – Dodo
    Commented Jan 30 at 11:48

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