calculate the amount of (nh4)2 which must be added to 500 ml of 0.2 nh3 to yield a solution of ph 9.35

Answer is 10.494g but mine is coming 5.247

To determine the amount of ammonium sulfate, \((NH_4)_2SO_4\), needed to achieve a specific pH in a solution of ammonia, we can use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base. In this case, ammonia (\(NH_3\)) acts as the base, and ammonium ion (\(NH_4^+\)) acts as the acid.

Understanding the Components

First, let’s clarify the components involved:

• Ammonia (\(NH_3\)): A weak base with a given concentration.
• Ammonium ion (\(NH_4^+\)): The conjugate acid of ammonia, which we will add to the solution.
• pH: The measure of acidity or basicity of the solution, which we want to adjust to 9.35.

Key Constants

The pKa of the ammonium ion is approximately 9.25. This value is crucial for our calculations.

Applying the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is expressed as:

pH = pKa + log([A-]/[HA])

In this equation:

• [A-] is the concentration of the base (in this case, \(NH_3\)).
• [HA] is the concentration of the acid (in this case, \(NH_4^+\)).

Calculating the Required Ratio

We want to find the ratio of \([NH_4^+]\) to \([NH_3]\) that will give us a pH of 9.35:

Substituting the known values into the equation:

9.35 = 9.25 + log([NH_4^+]/[NH_3])

This simplifies to:

0.10 = log([NH_4^+]/[NH_3])

To eliminate the logarithm, we can exponentiate both sides:

[NH_4^+]/[NH_3] = 10^{0.10} ≈ 1.2589

Finding the Concentration of Ammonium Ion

Next, we need to calculate the concentration of \(NH_4^+\) required. We know the concentration of \(NH_3\) in the solution:

[NH_3] = 0.2 \, M

Using the ratio we found:

[NH_4^+] = 1.2589 \times [NH_3] = 1.2589 \times 0.2 \, M ≈ 0.25178 \, M

Calculating the Amount of Ammonium Sulfate Needed

Now, we need to find out how much \((NH_4)_2SO_4\) we need to add to achieve this concentration in 500 mL of solution:

Volume of solution = 0.5 \, L

Using the formula:

moles = concentration × volume

We find the moles of \(NH_4^+\) needed:

moles \, of \, NH_4^+ = 0.25178 \, M \times 0.5 \, L = 0.12589 \, moles

Converting Moles to Grams

The molar mass of \((NH_4)_2SO_4\) is approximately 132.14 g/mol. Therefore, the mass required is:

mass = moles × molar mass

mass = 0.12589 \, moles × 132.14 \, g/mol ≈ 16.67 \, g

Final Adjustment

It seems there was a misunderstanding in your calculations. The amount of ammonium sulfate needed to achieve a pH of 9.35 in your solution is approximately 16.67 grams, not 10.494 grams or 5.247 grams. Ensure that you carefully follow each step and check your calculations for accuracy.