Question

Sodium ammonium hydrogen phosphate, \(\mathrm{NaNH}_{4}\) \(\mathrm{HPO}_{4},\) is a salt in which one of the ionizable \(\mathrm{H}\) atoms of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is replaced by \(\mathrm{Na}^{+},\) another is replaced by \(\mathrm{NH}_{4}^{+},\) and the third remains in the anion \(\mathrm{HPO}_{4}^{2-}\) Calculate the \(\mathrm{pH}\) of \(0.100 \mathrm{M} \mathrm{NaNH}_{4} \mathrm{HPO}_{4}(\mathrm{aq})\) [Hint: You can use the general method introduced on page \(720 .\) First, identify all the species that could be present and the equilibria involving these species. Then identify the two equilibrium expressions that will predominate and eliminate all the species whose concentrations are likely to be negligible. At that point, only a few algebraic manipulations are required.]

Short answer

The pH of the 0.100 M \(\mathrm{NaNH}_{4} \mathrm{HPO}_{4}\) solution is approximately 5.1.

Step-by-step solution

Identify the relevant species

There are three possible species that could be present in solution: the \(\mathrm{NH}_4^+,\) which could donate a proton (H+) to water; the \(\mathrm{Na}^+\) ion, which is a spectator ion; and \(\mathrm{HPO}_4^{2-}\), which may accept a proton from water. Since sodium ions (\(\mathrm{Na}^+\)) do not react with water and do not contribute to the pH, they can be disregarded.

Write out the equilibria

The equilibria in which \(\mathrm{NH}_4^+\) and \(\mathrm{HPO}_4^{2-}\) are involved can be written as follows: \[ \mathrm{NH}_4^+ + \mathrm{H}_2\mathrm{O} \leftrightarrows \mathrm{NH}_3 + \mathrm{H}_3\mathrm{O}^+\] \[ \mathrm{HPO}_4^{2-} + \mathrm{H}_2\mathrm{O} \leftrightarrows \mathrm{H}_2\mathrm{PO}_4^- + \mathrm{OH}^-\]

Consider the equilibrium expressions and constants

The equilibrium constant for the first reaction is represented by \(K_a\) (for the ammonium ion, \(\mathrm{NH}_4^+\)) and for the second one by \(K_b\) (for the phosphate ion, \(\mathrm{HPO}_4^{2-}\)). The constant \(K_a\) for \(\mathrm{NH}_4^+\) (\(K_a = 5.56 \times 10^{-10}\)) is significantly larger than the constant \(K_b\) for \(\mathrm{HPO}_4^{2-}\) (\(K_b = 2.00 \times 10^{-13}\)). Therefore, the equilibrium for the \(\mathrm{NH}_4^+\) reaction will contribute more to the [H+] in the solution and the second reaction can be neglected.

Calculate the \(H_3O^+\) concentration

The \(K_a\) expression for the first reaction is \[K_a=\frac{[\mathrm{NH}_3][\mathrm{H}_3\mathrm{O}^+]}{[\mathrm{NH}_4^+]}\] Write down the changes in concentration for each species in order for equilibrium to be achieved. We will assume \(\mathrm{NH}_4^+\) dissociates and loses a small concentration \(x\) which goes to \(\mathrm{NH}_3\) and \(\mathrm{H}_3\mathrm{O}^+\). Therefore, \[\mathrm{NH}_4^+\]; initial = 0.10 M, change = -x M, equilibrium = 0.10-x M. \[\mathrm{H}_3\mathrm{O}^+\] and \[\mathrm{NH}_3]\; initial = 0 M, change = +x M, equilibrium = x M. Substitute these into the \(K_a\) equation: \(5.56 \times 10^{-10}\) = \(\frac{x^2}{(0.10-x)}\) Since \(x<<0.10\), we can simplify to \(5.56 \times 10^{-10}\) = \(\frac{x^2}{(0.10)}\), which simplifies to \(x=\sqrt{(5.56 \times 10^{-10})(0.10)}\), giving \(x= 7.46 \times 10^{-6}\). This is the \(\mathrm{H}_3\mathrm{O}^+\) concentration at equilibrium.

Calculate the pH

Finally, use the pH equation, \[pH = -\log[\mathrm{H}^+]\] substituting the value of \(\mathrm{H}^+ = 7.46 \times 10^{-6}\) M, giving an final pH value of 5.1.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is fundamental in determining the pH level of a solution, especially when dealing with acid-base reactions. Chemical equilibrium occurs when the rates of the forward and reverse reactions in a closed system are equal, leading to constant concentrations of the reactants and products over time. This state can be disturbed by changes in concentration, pressure, or temperature, invoking Le Chatelier's Principle, which predicts the direction in which the reaction will move to re-establish equilibrium.

In the case of the dissociation of sodium ammonium hydrogen phosphate, equilibrium plays a pivotal role in finding the pH. Certain ions in the salt, specifically \(\text{NH}_4^+\) and \(\text{HPO}_4^{2-}\), engage in acid-base reactions with water, reaching a state of equilibrium. These equilibria are dynamic, meaning that even though the concentrations do not change, the reactions are still proceeding in both the forward and reverse directions.

Acid-Base Reactions

Acid-base reactions are processes where an acid donates a proton to a base. In the context of the given problem, \(\text{NH}_4^+\) acts as an acid when it donates a proton to water, forming \(\text{NH}_3\) (ammonia) and \(\text{H}_3\text{O}^+\) (hydronium ion). Conversely, \(\text{HPO}_4^{2-}\) can act as a base and accept a proton, yielding \(\text{H}_2\text{PO}_4^-\) and \(\text{OH}^-\) (hydroxide ion).

These reactions are reversible and crucial for determining the pH of the solution. The degree to which these reactions proceed is influenced by the strength of the acid or base, which is quantified by equilibrium constants (\(K_a\) and \(K_b\)). The pH calculation hinges on these reactions reaching equilibrium and the relative proportions of hydronium and hydroxide ions produced.

Equilibrium Constants

The equilibrium constants, \(K_a\) and \(K_b\), quantify the extent of acid or base dissociation in water at equilibrium. These values allow chemists to predict the position of equilibrium and, consequently, the concentrations of different ionic species in a solution.

In the given exercise, \(K_a\) for the ammonium ion (\(\text{NH}_4^+\)) is much larger than \(K_b\) for the phosphate ion (\(\text{HPO}_4^{2-}\)), indicating that the dissociation of \(\text{NH}_4^+\) will have a greater effect on the solution's acidity. By knowing the initial concentration of \(\text{NH}_4^+\) and \(K_a\), we can use the ICE table (Initial, Change, Equilibrium) method to calculate the concentration of \(\text{H}_3\text{O}^+\) at equilibrium, which is then used to determine the solution's pH.

Salt Hydrolysis

Salt hydrolysis is the reaction of salt ions with water, resulting in the formation of the weak acid or base and a modification of the solution's pH. This process is a specific type of acid-base reaction where the salt, being the product of an acid-base neutralization reaction, interacts with water to form the acid and base from which it was derived.

In our exercise, sodium ammonium hydrogen phosphate undergoes hydrolysis and \(\text{NH}_4^+\) reacts with water to produce \(\text{NH}_3\) and \(\text{H}_3\text{O}^+\), whereas \(\text{HPO}_4^{2-}\) reacts lesser due to its lower \(K_b\). The acid generated in these reactions leads to a decrease in the pH of the solution. By ignoring the contribution of \(\text{Na}^+\) which does not undergo hydrolysis, and focusing on the predominant reaction (in this case, the hydrolysis of \(\text{NH}_4^+\)), we can simplify the pH calculation.