Acid-Base Properties of Salts
The baking soda in your kitchen is a salt, and so is table salt. Yet a baking soda solution is clearly basic (that's why it works as an antacid), while a table salt solution is neutral. Both dissolve into ions. What makes the difference is what those ions do once they are free in the water.
Why a Salt Solution Can Be Non-Neutral
Suppose you measure the pH of three $0.10\ \text{M}$ solutions in the lab: $\mathrm{NaCl}$, $\mathrm{NH_4Cl}$, and $\mathrm{CH_3COONa}$. A pH meter reads $7.0$, $5.1$, and $8.9$. None of these salts contains $\mathrm{H^+}$ or $\mathrm{OH^-}$ to begin with. So why do two of them change the acidity of water?
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.
Cations of Weak Bases: Acidic Solutions
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.
Anions of Weak Acids: Basic Solutions
An aqueous solution is prepared by dissolving $4.10\text{ g}$ of sodium acetate, $\mathrm{CH_3COONa}$, in distilled water to a total volume of $500\text{ mL}$ at…
An aqueous solution is prepared by dissolving $4.10\text{ g}$ of sodium acetate, $\mathrm{CH_3COONa}$, in distilled water to a total volume of $500\text{ mL}$ at $25\,^{\circ}\mathrm{C}$.
- a) Write the equation for the chemical hydrolysis equilibrium and determine the value of the hydrolysis constant $K_\mathrm{h}$.
- b) Calculate the molar concentration of all the species present at equilibrium ($\mathrm{Na^+}$, $\mathrm{CH_3COO^-}$, $\mathrm{CH_3COOH}$, $\mathrm{OH^-}$ and $\mathrm{H_3O^+}$) and determine the final pH of the solution.
Data at $25\,^{\circ}\mathrm{C}$: Atomic molar masses in $\text{g}\times\text{mol}^{-1}$: $\mathrm{C}=12.0$; $\mathrm{H}=1.0$; $\mathrm{O}=16.0$; $\mathrm{Na}=23.0$. $K_\mathrm{a}(\mathrm{CH_3COOH})=1.8\times 10^{-5}$; $K_\mathrm{w}=1.0\times 10^{-14}$. Consider the solutions to be ideal.
Salts of a Weak Acid and a Weak Base
Ammonium acetate gives a solution very close to neutral, while ammonium fluoride gives one that is slightly below $7$. When both ions can react with water, what decides the result?
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, $\mathrm{NaHCO_3}$, gives $\mathrm{Na^+}$ (spectator) and $\mathrm{HCO_3^-}$. This anion acts as a base ($\mathrm{HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-}$, $K_b = K_w/K_{a1} \approx 2.3\times 10^{-8}$), and that tendency beats its acid behavior ($K_{a2} = 4.8\times 10^{-11}$). So the solution is basic, with $\mathrm{pH} \approx 8.3$.