Finding the pH of the solution

Question

In this video, at 3:30 min professor solves two equations $$\ce{H2O —> H+ + OH-}$$ and $$ \ce{HCl —> H+ + Cl-}$$

and deduces $[\ce{H+}] = [\ce{Cl-}] +[\ce{OH-}]$. How come this happen from those two equations?

Can anyone elaborate and explain?

Answer 1

For a quick general answer, this equation represents the charge balance. In a solution, the charges of all ions should add up to zero. You charge balance reads:

$$\ce{[H+] = [Cl-] + [OH-]}$$

If you had divalent ions, you would have to correct for that, e.g. if the solution also contained magnesium ions:

$$\ce{[H+] + 2 [Mg^2+] = [Cl-] + [OH-]}$$

How come this happen from those two equations?

$$\ce{H2O —> H+ + OH- }\tag{1}$$ $$\ce{HCl —> H+ + Cl- }\tag{2}$$

Both equations are balanced, which includes charge balance. For every $\ce{H+}$ made by dissociation of water, an $\ce{OH-}$ is also made. For every $\ce{H+}$ made by dissociation of hydrogen chloride, an $\ce{Cl-}$ is also made. So the total $\ce{H+}$ concentration is equal to the sum of the $\ce{OH-}$ and $\ce{Cl-}$ concentrations.

[OP's comment] Won't it be HCl+H2O =2 [H+]+ [OH-] + [Cl-] ?

This is a bit a step in the wrong direction, but I'll explain it anyway.

It looks like you are trying to add the two reactions. This would be appropriate if they were coupled, i.e. if a nanomachine grabs one molecule of hydrogen chloride and one molecule of water, and spews out two protons, a hydroxid and a chloride. The overall equation of that would be:

$$\ce{HCl +H2O ->2 H+ + OH- + Cl-}$$

For example, one mole HCl and one mole water would make two moles $\ce{H+}$, one mole hydroxide and one mole chloride. This would maintain the charge balance.

However, these two reactions are typically not coupled but happen independently. In fact, usually when reaction (2) increases the concentration of $\ce{H+}$, reaction (1) would go in reverse, lowering the hydroxide concentration (and the $\ce{H+}$ concentration).

Deriving the charge balance from the two balanced equations

The two reactions are independent in the sense that there is no set stoichiometric factor linking them. So let's say reaction (1) makes a certain amount $x$ of $\ce{H+}$, and reaction (2) makes a certain amount $y$. If we start out with no ions at all, and ions are only produced by these two reactions, we can write down all the amounts:

$$n_{\ce{H+},\mathrm{total}} = x + y$$

$$n_\ce{OH-} = x$$

$$n_\ce{Cl-} = y$$

Adding up the anions:

$$ n_\ce{OH-} + n_\ce{Cl-} = x + y $$

So

$$n_{\ce{H+},\mathrm{total}} = n_\ce{OH-} + n_\ce{Cl-}$$

To turn this into a statement about concentrations, divide the amounts by the volume they are in. Because it is the same volume for all (they are all in the same solution), we get a similar equation for the concentrations:

$$[\ce{H+}]_\mathrm{total} = [\ce{OH-}] + [\ce{Cl-}]$$

Example with numbers

For example, if reaction (1) made $\pu{e-10 M}$ of $\ce{H+}$, and reaction (1) made $\pu{e-4 M}$ of $\ce{H+}$, there would also be $\pu{e-10 M}$ of $\ce{OH-}$ and $\pu{e-4 M}$ of $\ce{Cl-}$.

Answer 2

Let's take the degree of dissociation of water be $\alpha$ and the $[\ce{H2O}] = c$, then at equilibrium the $[\ce{H^+}] = [\ce{OH^-}] = c\alpha$. And since $\ce{HCl}$ is a strong acid it will completely dissociate. So $\ce{HCl}$ is a then the $[\ce{H^+}] = [\ce{Cl^-}] = a$. So $[\ce{H^+}]_{\textrm{net}} = a+ c\alpha$, which is equal to $[\ce{OH^-}] + [\ce{Cl^-}]$.