Calculate the $\mathrm{pH}$ of a $\pu{1.5e–11 M}$ solution of $\ce{HCl}.$ How is the $\mathrm{pH}~7?$


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    $\begingroup$ Whats wrong with the pH being 7? The pH would actually be slightly lower than 7 but I'm assuming it will round to 7.00 as it is standard to give pH to 2 dp. $\endgroup$ – H.Linkhorn Jun 23 at 17:10
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    $\begingroup$ What pH would you expect, considering water self ionisation ? $\endgroup$ – Poutnik Jun 23 at 20:20

Pure water dissociate according to the equation: $$\ce{2H2O(l) <=> H3O+(aq) + OH- (aq)}$$

Assume that $K_\mathrm{w} = 1.00 \times 10^{-14}$ at room temperature. Thus, $\ce{[H3O+]}$ of solution is $\pu{1.00E-7 M }$. Now, if you add a trace amount of strong acid, this equilibrium would be disturbed and backward reaction occur to reduced some of added acid, according to the Le Chatelier's principle.

Suppose you added $\pu{1.5E−11 mol}$ of $\ce{HCl}$ to $\pu{1.0 L}$ of pure water (this means you made $\pu{1.5E−11 M}$ $\ce{HCl}$ solution). Since $\ce{HCl}$ strong acid, it would completely dissociate to $\ce{H3O+}$ and $\ce{Cl-}$. As stated above, the dissociation of water is vary depending on $\ce{[H3O+]}$ concentration of the solution. If $\alpha$ amount of water dissociates due to addition of $\ce{HCl}$, the total concentration of the $\ce{H3O+}$ in the solution will be $1.5\times 10^{−11} + \alpha$ and the concentration of the $\ce{OH-}$ in the solution will be $\alpha$. $$\therefore \; K_\mathrm{w} = [\ce{OH-}][\ce{H3O+}]= (1.5\times 10^{−11} + \alpha)\alpha = \alpha^2 + 1.5\times 10^{−11}\alpha = 1.00 \times 10^{-14}$$ Using the solution for quadratic equation,

$$\alpha= \frac{-1.5\times10^{-11} \pm \sqrt{(-1.5\times10^{-11})^2+ 4.00\times 10^{-14}}}{2} \approx \frac{-1.5\times10^{-11} \pm 2.00\times 10^{-7}}{2} \\\approx 1.00\times 10^{-7}$$

Thus, the total concentration of the $\ce{H3O+}$ in the solution is $(1.5\times 10^{−11} + 1.00\times 10^{-7}) \approx 1.00\times 10^{-7}$.

Therefore pH of the solution is 7.00.


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