Question

A buffer is prepared by adding $3.5\mathrm{g}$ of ammonium chloride $\left({\mathrm{NH}}_4\mathrm{Cl}\right)$ to $100\mathrm{mL}$ of $1.00\mathrm{M}{\mathrm{NH}}_3$ solution. (a) What is the $\mathrm{pH}$ of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

Short answer

a) The pH of the buffer is 9.43. b) The complete ionic equation for the reaction with hydrochloric acid added is: $N{H}_3+H^{+}+C{l}^{-}\to N{H}_4^{+}+C{l}^{-}$ c) The complete ionic equation for the reaction with sodium hydroxide added is: $N{H}_4^{+}+O{H}^{-}\to N{H}_3+H_{2}O$

Step-by-step solution

Calculate the concentration of NH4+ and NH3 in the buffer

First, we need to find the concentration of ammonium ions and ammonia in the buffer. We are given 3.5g of ammonium chloride (NH4Cl), which is added to 100 mL of 1.00 M NH3 solution. To find the concentration of ammonium ions, we need to convert grams of NH4Cl to moles: Moles of NH4Cl = (3.5 g) / (Molecular Weight of NH4Cl) Molecular Weight of NH4Cl = 14 (N) + (1 * 4) (H) + 35.5 (Cl) = 53.5 g/mol Moles of NH4Cl = (3.5 g) / (53.5 g/mol) = 0.0654 moles Concentration of NH4+ = moles / volume (in liters) = 0.0654 moles / 0.100 L = 0.654 M Since NH4Cl is added to the solution containing NH3, the concentration of NH3 remains 1.00 M.

Calculate the pH of the buffer

Now we can use the Henderson-Hasselbalch equation to find the pH of the buffer: pH = pKa + log([A-]/[HA]) For NH3/NH4+ buffer, pKa = 9.25 pH = 9.25 + log([NH3] / [NH4+]) pH = 9.25 + log(1.00 / 0.654) pH = 9.25 + 0.183 pH = 9.43 So, the pH of the buffer is 9.43.

Complete ionic equation with hydrochloric acid (HCl) added

Now we will write the complete ionic equation when hydrochloric acid (HCl) is added to the buffer. The balanced reaction is: NH3 + HCl → NH4+ + Cl- Now, we will represent the soluble compounds in their ionic form: NH3 + H+ + Cl- → NH4+ + Cl-

Complete ionic equation with sodium hydroxide (NaOH) added

Now we write the complete ionic equation when sodium hydroxide (NaOH) is added to the buffer. The balanced reaction is: NH4+ + OH- → NH3 + H2O Now, we will represent the soluble compounds in their ionic form: NH4+ + OH- → NH3 + H2O So, the final answers are: a) The pH of the buffer is 9.43. b) The complete ionic equation for the reaction with hydrochloric acid added is: NH3 + H+ + Cl- → NH4+ + Cl- c) The complete ionic equation for the reaction with sodium hydroxide added is: NH4+ + OH- → NH3 + H2O

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an incredibly useful tool when dealing with buffer solutions. A buffer solution maintains a relatively constant pH level even when acids or bases are added. It contains a weak acid and its conjugate base, or a weak base and its conjugate acid. This equation allows us to calculate the pH of buffer solutions effectively.
For our ammonium chloride and ammonia buffer system, the equation is expressed as:

• $extpH=extp{K}_a+extlog\left(\frac{[extBase]}{[extAcid]}\right)$

The $extp{K}_a$ is the negative log of the acid dissociation constant, which reflects the ability of the acid to donate protons. For the $ext{NH}_3/ext{NH}_4^{+}$ system, $extp{K}_a$ is 9.25.
The ratio of the concentrations of ammonia $[ext{NH}_3]$ to ammonium $[ext{NH}_4^{+}]$ ions in the buffer is crucial in determining the pH. In our example, ammonia is present at 1.00 M, and ammonium ions at 0.654 M. By applying these values to the equation, we calculate the buffer's pH as 9.43. This demonstrates how the buffer resists changes in pH upon additions of small amounts of acids or bases.

pH Calculation

Calculating the pH of a buffer solution involves understanding the relation between the concentrations of the buffer components. The pH indicates how acidic or basic a solution is. Values below 7 are acidic, while values above 7 are basic, and 7 is neutral.
To find the pH:

• Identify the weak acid and its conjugate base, or vice versa.
• Use the formula $extpH=extp{K}_a+extlog\left(\frac{[extBase]}{[extAcid]}\right)$
• Substitute the known concentrations and the $extp{K}_a$ value.

In our example, we followed these steps with ammonia $[ext{NH}_3]=1.00\text{M}$ as the base and ammonium $[ext{NH}_4^{+}]=0.654\text{M}$ as the conjugate acid. By substituting these values into the Henderson-Hasselbalch equation, we determined the pH to be 9.43.
This calculation helps us understand the buffer's pH and adjust formulations in lab settings if necessary. Regular use of buffer solutions is crucial in chemistry and biology as they maintain stable pH which is essential for many reactions.

Ionic Equations

Ionic equations are a way to show the ions involved in a chemical reaction. They are particularly useful for visualizing reactions in aqueous solutions where the dissociation of ionic compounds occurs.
In the case of our buffer solution, we observe the following:

• When a small amount of HCl is added to the buffer, the ammonium ion $ext{NH}_4^{+}$ neutralizes the incoming H${}^{+}$ ions from hydrochloric acid, forming ammonium again. The complete ionic equation can be written as: $ext{NH}_3+ext{H}^{+}+ext{Cl}^{-}\to ext{NH}_4^{+}+ext{Cl}^{-}$.
• When NaOH is added, the hydroxide $ext{OH}^{-}$ ions are neutralized by the ammonium ion $ext{NH}_4^{+}$, producing water and ammonia. The ionic equation is: $ext{NH}_4^{+}+ext{OH}^{-}\to ext{NH}_3+ext{H}_{2}extO$.

Understanding these equations is essential for predicting how a buffer will respond to different conditions. By retaining the equilibrium, the buffer ensures the pH remains relatively constant, which is a foundational principle in analytical and synthetic chemistry.