Baking soda from your kitchen cabinet is a salt, just like table salt. Yet a baking soda solution is clearly basic (it is the reason people take it for heartburn), while a table salt solution is neutral. Both salts dissolve in water and release ions. What makes the difference is what those ions do once they are free.
Why a salt can be non-neutral
In the lab, a pH meter reads $7.0$, $5.1$ and $8.9$ for $0.10\text{ M}$ solutions of $\mathrm{NaCl}$, $\mathrm{NH_4Cl}$ and $\mathrm{CH_3COONa}$. None of these salts contains $\mathrm{H^+}$ or $\mathrm{OH^-}$, so why do two of them change the pH of water?
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Two flasks containing aqueous solutions of the same concentration are available in the lab: one of sodium chloride, $\mathrm{NaCl}$, and the other of potassium nitrite,…
Two flasks containing aqueous solutions of the same concentration are available in the lab: 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.
Why does this work? $\mathrm{NH_4^+}$ and $\mathrm{NH_3}$ are a conjugate pair. The weaker the base, the stronger its conjugate acid, so the product $K_a \cdot K_b$ always equals $K_\mathrm{w}$.
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Ammonium nitrate, $\mathrm{NH_4NO_3}$, is a compound widely used in fertilizer solutions at $25\,^{\circ}\mathrm{C}$. a) Write the ionic dissociation and hydrolysis…
Ammonium nitrate, $\mathrm{NH_4NO_3}$, is a compound widely used in fertilizer solutions at $25\,^{\circ}\mathrm{C}$.
a) Write the ionic dissociation and hydrolysis processes in water corresponding to this salt, indicating which species act as a Brønsted-Lowry acid or base.
b) Deduce, with justification, whether the pH of the resulting aqueous solution will be lower than, equal to or higher than $7$. Data at $25\,^{\circ}\mathrm{C}$: $K_\mathrm{b}(\mathrm{NH_3})=1.8\times 10^{-5}$. Consider the solutions to be ideal.
Ammonium acetate gives a solution very close to $\mathrm{pH} = 7$, while ammonium fluoride is noticeably acidic. When both ions can react with water, what decides the result?
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An aqueous solution of ammonium fluoride, $\mathrm{NH_4F}$, is prepared at $25\,^{\circ}\mathrm{C}$. Consider the following equilibrium constant data at that…
An aqueous solution of ammonium fluoride, $\mathrm{NH_4F}$, is prepared at $25\,^{\circ}\mathrm{C}$. Consider the following equilibrium constant data at that temperature: $K_\mathrm{a}(\mathrm{HF})=6.6\times 10^{-4}$, $K_\mathrm{b}(\mathrm{NH_3})=1.8\times 10^{-5}$ and $K_\mathrm{w}=1.0\times 10^{-14}$.
a) Write the ionic reactions that take place when both ions from the salt interact with water.
b) Compare quantitatively the constants of the hydrolysis equilibria of the ions present and determine whether the resulting solution is acidic, basic or neutral. Consider the solutions to be ideal.
In the general chemistry lab, four beakers numbered 1 to 4 have been labeled, each containing a $0.10\text{ M}$ aqueous solution at $25\,^{\circ}\mathrm{C}$ of the…
In the general chemistry lab, four beakers numbered 1 to 4 have been labeled, 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 setup 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.
Back to the start: Baking soda ($\mathrm{NaHCO_3}$) releases $\mathrm{Na^+}$ (a spectator) and $\mathrm{HCO_3^-}$. The basic hydrolysis of this anion ($\mathrm{HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-}$, with $K_b = K_\mathrm{w}/K_{a1} \approx 2.3\times 10^{-8}$) beats its acid dissociation ($K_{a2} = 4.8\times 10^{-11}$), so the solution is clearly basic, with $\mathrm{pH} \approx 8.3$.
Summary
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