What mass of ammonium nitrate must be added to 350 mL OF 0.150 M solution of ammonia to give a buffer having pH of 9.00? (Kb(NH3)= 1.8X10-5

To calculate the mass of ammonium nitrate required, we need to follow these steps:

Step 1: Write the balanced chemical equation for the reaction between ammonia (NH3) and ammonium nitrate (NH4NO3).
NH3 + H2O ⇌ NH4+ + OH-

Step 2: Calculate the concentration of hydroxide ions (OH-) in the buffer solution.
Since the pH of the buffer solution is 9.00, the concentration of hydroxide ions can be calculated using the equation: [OH-] = 10^(-pOH)
Given that pOH = 14 - pH = 14 - 9 = 5, [OH-] = 10^(-5) M

Step 3: Calculate the concentration of ammonia (NH3) using the equation:
[OH-] = Kb * [NH3] / [NH4+]
Given that Kb(NH3) = 1.8 x 10^(-5) M and that the concentration of ammonia (NH3) is initially 0.150 M, we can rearrange the equation to solve for [NH3]:
0.150 * 1.8 x 10^(-5) / [NH4+] = 10^(-5)
[NH4+] = (0.150 * 1.8 x 10^(-5)) / 10^(-5)
[NH4+] = 0.270 M

Step 4: Calculate the moles of ammonium ion (NH4+) present using the equation: moles = concentration x volume
Given that the volume of the solution is 350 mL = 0.350 L, we can calculate the moles of NH4+:
moles = 0.270 M x 0.350 L
moles = 0.0945 mol

Step 5: Calculate the molecular weight of ammonium nitrate (NH4NO3).
The molecular weight of NH4NO3 = (1 x 14.01) + (4 x 1.01) + (1 x 14.01) + (3 x 16.00)
The molecular weight of NH4NO3 = 80.04 g/mol

Step 6: Calculate the mass of ammonium nitrate required using the equation:
mass = moles x molecular weight
mass = 0.0945 mol x 80.04 g/mol
mass ≈ 7.56 grams

Therefore, approximately 7.56 grams of ammonium nitrate must be added to 350 mL of the 0.150 M ammonia solution to create a buffer with a pH of 9.00.

To solve this problem, we need to determine the amount of ammonium nitrate (NH4NO3) needed to achieve a buffer solution with a pH of 9.00.

First, we need to understand the reaction that occurs between ammonia (NH3) and water (H2O). Ammonia reacts with water to form ammonium ions (NH4+) and hydroxide ions (OH-).

The balanced equation for this reaction is:

NH3 + H2O ⇌ NH4+ + OH-

Since we are working with a buffer solution, we want to maintain a specific pH. The pH is determined by the concentration of hydroxide ions (OH-) in the solution. In this case, the pH is 9.00, which means the concentration of hydroxide ions is 10^(-5) M.

The Kb value for ammonia (NH3) is given as 1.8x10^(-5).

The Kb expression for the reaction is:

Kb = [NH4+][OH-] / [NH3]

We know the concentration of the hydroxide ions and the Kb value, so we can rearrange the equation to solve for the concentration of NH3 (ammonia):

[NH3] = ([NH4+][OH-]) / Kb

= (x)(10^(-5)) / (1.8x10^(-5))

= x / 1.8

Now, we calculate the amount of NH3 required to make the buffer solution. The volume of the buffer is given as 350 mL, and the concentration of NH3 is given as 0.150 M.

moles of NH3 = (concentration of NH3) x (volume of solution)

= 0.150 M x 0.350 L

= 0.0525 moles

Thus, we need 0.0525 moles of ammonia (NH3) to achieve a buffer solution with a pH of 9.00.

Finally, we need to convert the moles of NH3 to grams of ammonium nitrate (NH4NO3). The molar mass of NH4NO3 is 80.04 g/mol.

mass of NH4NO3 = (moles of NH3) x (molar mass of NH4NO3)

= 0.0525 mol x 80.04 g/mol

= 4.203 g

Therefore, you would need to add approximately 4.203 grams of ammonium nitrate to 350 mL of the 0.150 M ammonia solution in order to achieve a buffer solution with a pH of 9.00.

pH = pKa + log (base)/(acid)

9.00 = pKa + log b/a
You get pKa from Kb. Convert Kb to pKb; then pKa + pKb = 14 and solve for pKa.
Substitute (base) from the problem and solve for (acid), then convert (acid) to grams NH4NO3. Post your work if you get stuck.

I got 2.36g. Is that correct?