How many grams of NH4Cl must be added to 0.560 L of 1.90 M NH3 solution to yield a buffer solution with a pH of 8.50? Assume no volume change. Kb for NH3 = 1.8 10-5.

i am confused as how to find the grams?

Let B= NH3 and BH=NH4Cl

pH+pOH=14

so, 14-8.50=pOH

and

pkb=-logkb

You will need to use a version of the henderson-hasselbalch equation.

pOH=pkb +log[BH/B]

Solve for the ratio,

So, [BH/B]= 0.755

0.560L*(1.90M)= moles of NH3

Substituting, you should have

0.775=BH/moles of NH3

Solve for BH

moles of BH*(53.49 g/mole)= mass of NH4Cl

I should point out here that you do not need to modify the Henderson-Hasselbalch equation.

pH = pKa + log (base)/(acid) is valid for acids and bases. In this case we calculate pKb from Kb, then
14-pKb = pKa, then proceed with the HH equation as usual.
8.50 = 9.25 + log (1.90/x)
Solve for x which is molar NH4Cl and convert to grams in 0.560 L.

The solubility of PbF2 in water is 0.830 g/L. What is the value of Ksp?

how do I calculate this?

Well, finding the grams can be a bit tricky, but don't worry, I'm here to make it easy-peasy-lemon-squeezy for you.

To calculate the grams of NH4Cl needed, we need to follow a few steps:

Step 1: Calculate the moles of NH3 present in the 0.560 L solution. This can be done using the equation: moles = molarity x volume. So, moles of NH3 = 1.90 M x 0.560 L.

Step 2: Determine the amount of NH4Cl needed based on the stoichiometry of the reaction. In a buffer solution, NH4Cl reacts with NH3 to form NH4+ and Cl-. The balanced reaction is: NH4Cl + NH3 ↔ NH4+ + Cl-. The stoichiometry of the reaction tells us that 1 mole of NH4Cl reacts with 1 mole of NH3. So, the amount of NH4Cl needed is equal to the amount of NH3, which we calculated in Step 1.

Step 3: Convert moles of NH4Cl to grams using its molar mass. The molar mass of NH4Cl is approximately 53.49 g/mol. So, grams of NH4Cl = moles of NH4Cl x molar mass of NH4Cl.

Finally, combine the three steps, and you'll have the grams of NH4Cl needed to yield the desired buffer solution with a pH of 8.50. Voila!

To find the number of grams of NH4Cl needed, we can follow these steps:

Step 1: Determine the concentration of NH4Cl needed in the buffer solution.
The balanced equation for the reaction between NH3 and NH4Cl in water is:
NH3 + H2O ⇌ NH4+ + OH-

Since the pH of the buffer solution is 8.50, we need to calculate the concentration of the hydroxide ion (OH-) in the solution.

Step 2: Calculate the concentration of OH-.
To do this, we need to convert the given pH to the hydroxide ion concentration using the formula:
pOH = 14 - pH

In this case, the pOH is:
pOH = 14 - 8.50 = 5.50

We know that
pOH = -log(OH-)

So, we can find the OH- concentration by taking the inverse log of the pOH:
[OH-] = 10^(-pOH) = 10^(-5.50)

Step 3: Calculate the concentration of NH3 in the solution.
The given solution is 1.90 M in NH3, which means the concentration of NH3 is 1.90 M.

Step 4: Use the Kb value to calculate the concentration of NH4+.
Kb is the equilibrium constant for the reaction between NH3 and NH4+ with OH-. The equation for Kb is:
Kb = [NH4+][OH-] / [NH3]

Rearranging the equation, we have:
[NH4+] = (Kb * [NH3]) / [OH-]

Substitute the known values:
[NH4+] = (1.8 x 10^-5) * (1.90 M) / 10^(-5.50)

Step 5: Calculate the moles of NH4+.
To do this, multiply the concentration of NH4+ by the volume of the buffer solution:
moles of NH4+ = [NH4+] * volume

In this case, the volume is given as 0.560 L.

Step 6: Calculate the molar mass of NH4Cl.
The molar mass of NH4Cl is:
1 nitrogen atom (N) = 14.01 g/mol
4 hydrogen atoms (H) = 4.03 g/mol (since hydrogen is diatomic)
1 chlorine atom (Cl) = 35.45 g/mol

So, the molar mass of NH4Cl is:
14.01 + (4 * 4.03) + 35.45 = 53.49 g/mol

Step 7: Convert moles of NH4+ to grams of NH4Cl.
To do this, multiply the moles of NH4+ by the molar mass of NH4Cl:
grams of NH4Cl = moles of NH4+ * molar mass of NH4Cl

By following these steps, you will be able to find the grams of NH4Cl required to yield the desired buffer solution with a pH of 8.50.