Problem 6 - Chemistry and Chemical Reactivity

Question

What is the pH of the solution that results from adding \(25.0 \mathrm{mL}\) of \(0.12 \mathrm{M} \mathrm{HCl}\) to \(25.0 \mathrm{mL}\) of \(0.43 \mathrm{M} \mathrm{NH}_{3} ?\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 9.66.

Step 1: Determine Moles of HCl

To find the moles of HCl, we use the formula \( ext{moles} = ext{concentration} \times ext{volume} \). Here, the volume needs to be in liters, so we convert it: \( 25.0 \, \text{mL} = 0.0250 \, \text{L} \). The moles of HCl are then \( 0.12 \, \text{mol/L} \times 0.025 \, \text{L} = 0.0030 \, \text{mol} \).

Step 2: Determine Moles of NH extsubscript{3}

Similarly, calculate the moles of \(\text{NH}_3\) using its concentration and volume (also converted to liters). \( 0.43 \, \text{mol/L} \times 0.025 \, \text{L} = 0.01075 \, \text{mol} \).

Step 3: Write the Neutralization Reaction

When HCl is added to \(\text{NH}_3\), the reaction is \(\text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4^+ + \text{Cl}^-\). This is a neutralization reaction where the acid (HCl) reacts with the base (\(\text{NH}_3\)).

Step 4: Determine Limiting Reactant

Compare the moles of HCl and \(\text{NH}_3\). Since \(0.0030 \, \text{mol HCl} < 0.01075 \, \text{mol \(\text{NH}_3\)}\), HCl is the limiting reactant. It will react completely with an equivalent amount of \(\text{NH}_3\), leaving \(0.01075 - 0.0030 = 0.00775 \, \text{mol \(\text{NH}_3\)}\) unreacted.

Step 5: Calculate Concentration of Remaining NH extsubscript{3} and NH extsubscript{4} extsuperscript{+}

The concentration of \(\text{NH}_3\) that remains is given by \( \frac{0.00775 \, \text{mol}}{0.050 \, \text{L}} = 0.155 \, \text{M} \). The concentration of \(\text{NH}_4^+\), which is equal to the initial moles of \(\text{HCl}\), is \(\frac{0.0030 \, \text{mol}}{0.050 \, \text{L}} = 0.060 \, \text{M}\).

Step 6: Calculate pH Using Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation for a buffer solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{NH}_3]}{[\text{NH}_4^+]} \right) \). First, find \(\text{pK}_a\) from \(\text{pK}_w = 14\) and \(\text{pK}_b\) of \(\text{NH}_3\) (\(\approx 4.75\)): \(\text{pK}_a = 14 - 4.75 = 9.25\). The ratio \( \frac{[\text{NH}_3]}{[\text{NH}_4^+]} \) is \( \frac{0.155}{0.060} \approx 2.583\). Thus, \( \text{pH} = 9.25 + \log(2.583) \).

Step 7: Calculate Final pH

Compute the logarithmic term \( \log(2.583) \approx 0.412 \) and add it to \(\text{pK}_a\) to get the final pH: \(9.25 + 0.412 = 9.662\). Thus, the pH of the solution is approximately 9.66.

Key Concepts

Neutralization Reaction

In acid-base chemistry, a neutralization reaction is a fundamental process where an acid and a base react to form water and a salt. This reaction helps balance hydrogen ions (\(H^+\)) from the acid with hydroxide ions (\(OH^-\)) from the base, leading to a neutral or near-neutral solution.
When hydrochloric acid (\(HCl\)) is mixed with ammonia (\(NH_3\)), a neutralization occurs:

• \(HCl + NH_3 \rightarrow NH_4^+ + Cl^-\)

Here, the acid (\(HCl\)) donates protons to form ammonium ions (\(NH_4^+\)), and the chloride ions (\(Cl^-\)) remain in solution.
This reaction is key in many areas of chemistry, including buffer creation and salt formation.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simple mathematical formula used to estimate the pH of a buffer solution. It relates the pH, pKa of the acid, and the ratio of the concentrations of the conjugate base and acid:
\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]In this exercise, it helps calculate the pH after HCl and \(NH_3\) have reacted.
First, we determine \(pK_a\) by subtracting the base's \(pK_b\) from 14 (since \(\text{pK}_w = 14\)): \(pK_a = 14 - 4.75 = 9.25\).

• The base \([NH_3]\) concentration is 0.155 M.
• The acid \([NH_4^+]\) concentration is 0.060 M.

The ratio used in the equation gives insight into the remaining concentrations post-reaction, leading to an accurate pH calculation of approximately 9.66.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is used up first, and it determines the amount of product formed. Identifying the limiting reactant is crucial for accurate calculations.
Here, we calculate the moles of \(HCl\) and \(NH_3\):

• \(0.0030 \, \text{mol}\) of \(HCl\)
• \(0.01075 \, \text{mol}\) of \(NH_3\)

Since \(0.0030 \, \text{mol}\) of \(HCl\) is less than \(0.01075 \, \text{mol}\) of \(NH_3\), \(HCl\) is the limiting reactant.
This means it will fully react with \(NH_3\), leaving some \(NH_3\) unreacted. Knowing the limiting reactant allows us to proceed with concentration and pH calculations correctly.

Acid-Base Chemistry

Acid-base chemistry involves the study of how acids and bases interact in a solution. Acids donate protons \( (H^+) \), while bases accept them, which leads to various reactions and pH changes.
For this exercise, understanding the properties of \(HCl\) as a strong acid and \(NH_3\) as a weak base is key. These interactions form the basis of calculating resulting concentrations and pH.
After the reaction:

• HCl completely dissociates, contributing \(H^+\) ions.
• NH3 reacts to form \(NH_4^+\), impacting the solution’s pH balance.

This knowledge is foundational to predicting the behavior of solutions and is widely applied in chemistry and biochemistry fields.