Bread soda (sodium hydrogencarbonate) from the kitchen press is a salt, and so is the table salt on your chips. Yet a solution of bread soda is clearly basic, which is why it was used to settle an upset stomach, while a solution of table salt is neutral. Both are salts and both dissolve in water to give ions. What makes the difference is what those ions do once they are free.
Why a salt may not be neutral
Measure 0.1 M solutions of $\mathrm{NaCl}$, $\mathrm{NH_4Cl}$ and $\mathrm{CH_3COONa}$ with a pH meter and you get about $7.0$, $5.1$ and $8.9$. None of these salts contains $\mathrm{H^+}$ or $\mathrm{OH^-}$ to begin with. So how can they change the acidity of pure water?
To predict a salt's behaviour, ask which acid and which base it came from. The strong ones you need to know are:
Why does this work? The Brønsted–Lowry model tells us that the conjugate of a weak acid is a base that holds on to protons reasonably well. The conjugate of a strong acid is a very weak base that does not take protons from water at all. For any conjugate pair:
Two flasks containing aqueous solutions of the same concentration are available in the laboratory: one of sodium chloride, $\mathrm{NaCl}$, and the other of potassium…
Two flasks containing aqueous solutions of the same concentration are available in the laboratory: one of sodium chloride, $\mathrm{NaCl}$, and the other of potassium nitrite, $\mathrm{KNO_2}$, at $25\,^{\circ}\mathrm{C}$.
a) Write the equations for the dissociation of both salts in water and identify the spectator ions that do not undergo hydrolysis.
b) Justify the acidic, basic or neutral character of each of the two resulting solutions, given that $K_\mathrm{a}(\mathrm{HNO_2})=4.5\times 10^{-4}$ at $25\,^{\circ}\mathrm{C}$. Consider the solutions to be ideal.
The equilibrium constant for a hydrolysis reaction is not a new kind of constant. It is just $K_a$ of the cation (or $K_b$ of the anion), found from $K_w$ using the conjugate pair. (Some books call it $K_h$, the hydrolysis constant, an extension beyond the syllabus.)
Because $K_a$ is tiny, very little of the ion reacts. We can then treat it like any weak acid, with $x = [\mathrm{H_3O^+}]$ and $[\mathrm{NH_4^+}] \approx c_0$:
The degree of hydrolysis is the fraction of the ion that has reacted: $\alpha_h = x / c_0$. This is the same idea as the degree of dissociation of a weak acid, applied to an ion.
Hydrolysis of anions: weak acid and strong base
The method mirrors the cation case, using $[\mathrm{OH^-}] \approx \sqrt{K_b\, c_0}$, then $\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}]$ and $\mathrm{pH} = 14 - \mathrm{pOH}$ at 25 °C. (Equivalently, $[\mathrm{H_3O^+}] = K_w/[\mathrm{OH^-}]$.)
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In the general chemistry laboratory, four beakers numbered 1 to 4 have been labelled, each containing a $0.10\text{ M}$ aqueous solution at $25\,^{\circ}\mathrm{C}$ of…
In the general chemistry laboratory, four beakers numbered 1 to 4 have been labelled, each containing a $0.10\text{ M}$ aqueous solution at $25\,^{\circ}\mathrm{C}$ of the following substances: $\mathrm{NaCl}$, $\mathrm{CH_3COONa}$, $\mathrm{NH_4Cl}$ and $\mathrm{KNO_3}$. The diagram of the experimental set-up is shown below:
a) Classify each of the four salts, indicating whether the acid and the base from which it derives are strong or weak, and state which undergo hydrolysis.
b) Given that $K_\mathrm{a}(\mathrm{CH_3COOH})=1.8\times 10^{-5}$, $K_\mathrm{b}(\mathrm{NH_3})=1.8\times 10^{-5}$ and $K_\mathrm{w}=1.0\times 10^{-14}$ at $25\,^{\circ}\mathrm{C}$, arrange the four solutions in order of increasing pH, justifying your answer. Consider the solutions to be ideal.
Take $0.10\ \text{M}$ $\mathrm{NH_4Cl}$ and dilute it ten times to $0.010\ \text{M}$. With $K_a = 5.6 \times 10^{-10}$ and $x = \sqrt{K_a c_0}$:
$c_0$ / M
$[\mathrm{H_3O^+}]$ / M
pH
$\alpha_h$
0.10
$7.5 \times 10^{-6}$
5.13
$7.5 \times 10^{-5}$
0.010
$2.4 \times 10^{-6}$
5.63
$2.4 \times 10^{-4}$
The pH moves closer to 7 (there is less $\mathrm{H_3O^+}$ in total), but a larger fraction of the ammonium ions has reacted. This is the same pattern as for a weak acid: $\alpha_h = x/c_0 = \sqrt{K/c_0}$, so diluting by a factor of 10 increases $\alpha_h$ by $\sqrt{10} \approx 3.2$. Why? On dilution every concentration in the equilibrium expression falls, so the reaction quotient $Q$ drops below $K_a$ and the equilibrium shifts towards the products until $Q = K_a$ again.
Equivalence point of a titration
At the equivalence point of a titration of a weak acid with a strong base, all the acid has been converted into its salt. The solution contains only that salt and water, so its pH is set by hydrolysis of the anion, and it is above 7. That is why the indicator for such a titration must change colour on the basic side.
Salts of a weak acid and a weak base
Ammonium ethanoate gives a solution of pH about 7.0, while ammonium fluoride gives about 6.2. When both ions can react with water, what decides the result?
For a salt $\mathrm{BH^+A^-}$, the overall equilibrium is $\mathrm{BH^+ + A^- \rightleftharpoons B + HA}$. To get the pH, write the two acid constants and multiply them:
The salt gives equal amounts of cation and anion, and the overall reaction produces equal amounts of B and HA. If hydrolysis is small, then $[\mathrm B] \approx [\mathrm{HA}]$ and $[\mathrm{BH^+}] \approx [\mathrm{A^-}]$, so the last fraction is 1. That gives
Notice that the concentration of the salt has cancelled out: the pH of this kind of salt does not depend on how concentrated it is.
Back to the start: bread soda, $\mathrm{NaHCO_3}$, releases $\mathrm{Na^+}$ (a spectator) and $\mathrm{HCO_3^-}$. This anion can act as a base ($\mathrm{HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-}$, with $K_b = K_w/K_{a1} \approx 2.3 \times 10^{-8}$) or as an acid (with $K_a = 4.8 \times 10^{-11}$). The base reaction is far stronger, so the solution is basic, with $\mathrm{pH} \approx 8.3$.
Summary
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