Asked earlier but got no reply :/ hopefully I get one now cause I'm really stuck and can't completely my assignment cause of one question..

How many grams of dry NH4Cl need to be added to 1.90 L of a 0.200 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.99? Kb for ammonia is1.8*10^-5.

Can anyone help me with this question? Thanks in advance!

nvm got it.

To answer your question, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentration of the acid and its conjugate base. The equation is:

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

In this case, ammonia (NH3) is the weak base and ammonium chloride (NH4Cl) is its conjugate acid. We want to prepare a buffer solution with a pH of 8.99. Since we know the pKa of ammonia, which is -log(Kb), we can calculate the pKa as follows:

pKa = -log(1.8 * 10^-5) = 4.74

Now that we have the pKa, we can rearrange the Henderson-Hasselbalch equation to solve for the ratio of [A-] to [HA]:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(8.99 - 4.74) = 7943

Since we have a 0.200 M solution of ammonia, we can assume the initial concentration of NH3 is 0.200 M, and the concentration of NH4+ will be much smaller than that. So we can approximate [A-] as 0.200 M.

Now we can calculate the concentration of NH4Cl needed in order to achieve a ratio of 7943:

[HA] = [A-] / 7943 = 0.200 M / 7943 ≈ 2.52 * 10^-5 M

To determine the mass of NH4Cl needed, we need to convert the concentration to moles and then to grams. First, calculate the number of moles:

moles of NH4Cl = (2.52 * 10^-5 M) * 1.90 L = 4.79 * 10^-5 mol

Finally, calculate the mass of NH4Cl using its molar mass:

mass of NH4Cl = moles of NH4Cl * molar mass
= (4.79 * 10^-5 mol) * (53.49 g/mol)
≈ 2.57 * 10^-3 g

Therefore, approximately 2.57 milligrams (mg) of dry NH4Cl need to be added to 1.90 L of the 0.200 M solution of ammonia to prepare the desired buffer solution with a pH of 8.99.