# Calculating the pH of Strong Acid Solutions [closed]

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

• 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. – H.Linkhorn Jun 23 '19 at 17:10
• What pH would you expect, considering water self ionisation ? – Poutnik Jun 23 '19 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 $$(1.5\times 10^{−11} + 1.00\times 10^{-7}) \approx 1.00\times 10^{-7}$$.