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I dissolved some sodium carbonate in water and I measured the $\mathrm{pH}$. It turned out to be 11. I don't really understand why dissolving it in water increased the $\mathrm{pH}$. I mean, I know that $\mathrm{pH}$ of water was 7, so it must have something to do with the $\ce{H+}$ ions. We just covered acid and bases today so please bear with me.

Here are my thoughts. My salt had some $\ce{OH-}$ salt and this combined with the $\ce{H+}$ to produce $\ce{H2O}$ thereby decreasing the amount of $\ce{H+}$ ion. Or maybe one of my salt ions combine with the dissolved $\ce{H+}$, again thereby decreasing the $\ce{H+}$ concentration and increasing the $\mathrm{pH}$, so I'd have something like this could have something like:

$$\ce{Na+ + H+ -> NaH2+}$$

or maybe

$$\ce{CO3^2- + H+ -> HCO3-}$$

Which one do you think happened?

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    $\begingroup$ Last one is OK. $\endgroup$
    – Mithoron
    Commented Jul 19, 2016 at 23:07
  • $\begingroup$ Could you tell me why? Is it just more likely? $\endgroup$
    – Hamze
    Commented Jul 19, 2016 at 23:34
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    $\begingroup$ This salt doesn't have any OH- and protonation of Na+ is nonsense in any solution. $\endgroup$
    – Mithoron
    Commented Jul 20, 2016 at 1:15

3 Answers 3

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It's a little more complicated than that.. When you dissolved $\ce{Na2CO3}$ in water you had a solution of ions: $\ce{Na+}$, $\ce{CO3^2-}$, $\ce{H3O+}$ and $\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7).

Now when you mix all those ions, the equilibrium isn't only the equilibrium of water dissociation ($\ce{H2O + H2O <=> H3O+ + OH-}$). Now you have a whole series of equilibria occurring in the solution simultaneously:

1) Water dissociation: $\ce{H2O + H2O <=> H3O+ + OH-}$

2) Salt dissociation: $\ce{Na2CO3 <=> 2Na+ + CO3^2-}$

3) Base equilibrium*: $\ce{NaOH <=> Na+ + OH-}$

4) Acid equilibrium*: $\ce{CO3^2- + 2 H3O+ <=> H2CO3 + 2 H2O}$

The salt dissociation doesn't affect the pH directly. We have two equilibria though that do unbalance the proportion of $\ce{H3O+}$ and $\ce{OH-}$ ions, which are the "base equilibrium" and "acid equilibrium". We don't think about them at first sight because we didn't add any $\ce{NaOH}$ or $\ce{H2CO3}$ to the solution, but everything needed to form them is right there.

$\ce{NaOH}$ is a strong base, however $\ce{H2CO3}$ is a weak acid. So it's likely (and you witnessed it in experience) that more $\ce{H3O+}$ will be consumed in comparison with the $\ce{OH-}$, making the solution more alkaline.

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  • $\begingroup$ Um, could you clarify on what exactly an equilibrium is? $\endgroup$
    – Hamze
    Commented Jul 19, 2016 at 23:30
  • $\begingroup$ I'm thinking of a simplified way of explaining it. A chemical reaction can have a direct path (of product formation) and a reverse path (where the products react to form the reactants again). Say we have $\ce{A+B <-> C}$. Just as A and B can react and form C, C also can react and form A and B again. But there is a tendency of each of them to happen and one of the variables that usually changes that tendency is the concentration of A, B and C. [continue...] $\endgroup$
    – IanC
    Commented Jul 19, 2016 at 23:43
  • $\begingroup$ Say you have 1M of A, 1M of B and 0M of C. The rate of $\ce{A +B -> C}$ will be much higher than the rate of $\ce{C -> A + B}$. But as time passes the concentration of C will increase, while the concentration of A and B will decrease. This will decrese the rate of $\ce{A+B -> C}$ and increase the rate of $\ce{C -> A + B}$, until the rates are the same. When this happens you will have reached the static equilibrium state of the reaction. In reactions like $\ce{NaOH <-> Na+ + OH-}$ equilibrium tend to the right ($\ce{Na+}$ and $\ce{OH-}$ are likely to be present in higher quantities than NaOH). $\endgroup$
    – IanC
    Commented Jul 19, 2016 at 23:48
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    $\begingroup$ @IanC I have feeling you're confusing OP even more. $\endgroup$
    – Mithoron
    Commented Jul 20, 2016 at 1:10
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    $\begingroup$ @Mithoron Well, I tried to simplify it as I could.. But it's hard to explain the reason behind salts acidity/basicity without mentioning chemical reactions/dissociation equilibrium. I was hoping the youtube video would clarify things a little for him $\endgroup$
    – IanC
    Commented Jul 20, 2016 at 1:31
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You have some detailed answers, but the principles are quite simple to grasp.

A 'strong' acid or base is one in which there is complete, i.e. 100% dissociation of the molecule when dissolved in water. Examples are $\ce{HCl}$ which dissociates into $\ce{H+}$ and $\ce{Cl-}$, and NaOH into $\ce{Na+}$ and $\ce{OH-}$. If 1 mole of each is added to pure water the solution remains neutral ($\mathrm{pH = 7}$) because there are equal numbers of $\ce{H+}$ and $\ce{OH-}$.

A 'weak' acid or 'weak' base is one which does not dissociate completely, so, for example, some fraction of $\ce{H2CO3}$ molecules remain in solution and the rest dissociate producing $\ce{H+}$ and $\ce{CO3^2-}$ ions. Because some protons remain bound in the $\ce{H2CO3}$ when 1 mole of strong base is added to 1 mole of the acid, there is an excess of $\ce{OH-}$ and the solution is alkaline. A similar argument applies to a weak base and strong acid.

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TL;DR: Because there is an equilibrium between anion of $\ce{CO3^2-}$ and $\ce{HCO3^-}$ that depletes $\ce{H+}$ and that causes more water to dissociate and form free $\ce{OH-}$.

You should start by reading about hydrolysis and buffers (chemistry). Briefly, you have these equilibria:

\begin{align} \ce{M+ + OH- &<=> MOH}& &\text{(limiting cases: $\ce{M+}$ = $\ce{Na+}$ or $\ce{NH4+}$)} \\ \ce{H+ + A- &<=> HA}& &\text{(limiting cases: $\ce{A-}$ = $\ce{I-}$ or $\ce{CH3COO-}$)} \\ \ce{H+ + OH- &<=> H2O} & \end{align}

Here $[\ce{Y}]$ means concentration of $\ce{Y}$. Note that $\frac{[\ce{MOH}]}{[\ce{M+}][\ce{OH-}]} = K_\mathrm{b}$, $\frac{[\ce{HA}]}{[\ce{H+}][\ce{A-}]} = K_\mathrm{a}$ and $[\ce{H+}][\ce{OH-}] = 10^{-14}$.

Depending on the $K_\mathrm{a}$ and $K_\mathrm{b}$ you have several situations. Acids and bases are called strong if they favor dissociation on ions and weak if they favor association of ions.

If both your acid and base is strong, then they both dissociate and the released $\ce{H+}$ and $\ce{OH-}$ are recombining to form water. If acid is strong and a base is weak then acid dissociates fully and yield a lot of $\ce{H+}$, but base doesn't dissociate fully and yield less $\ce{OH-}$. As a result of recombination of $\ce{H+}$ and $\ce{OH-}$, you get excess of $\ce{H+}$ and acidic condition. If the acid is weak and the base is strong you have more free $\ce{OH-}$ and basic reaction much by the same logic. If both acid and base are weak $\mathrm{pH}$ depends on relative strengths on acid and base.

In your case acid is weak ($\ce{CO2}$ forms slightly acidic solution in water) and base is strong ($\ce{NaOH}$ is a strong base). Thus, you have a basic solution.

This story is overly simplified because you have an acid with two stages of dissociation:

\begin{align} \ce{H2CO3 &<=> H+ + HCO3-} \\ \ce{HCO3- &<=> H+ + CO3^2-} \end{align}

This adds one equation to the system but similar logic applies.

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