How many grams of dry NH4Cl need to be added to 2.10L of a 0.600M solution of ammonia,NH3 , to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8x10^-5.

I just wanted to make sure I dd this write, can you please check my work?
Thank you!!!!

I found the pKa which was 9.255..
then I used that and plugged it in the Hasselbalch (i think that's what it's called) equation.
I got .92592....M for NH4Cl and i used that Molarity and multiplied it by the liters.
so the mass was 1.94 moles of NH4Cl
I then used that to find the grams and my answer came out to be 104.02 grams of NH4Cl

Would you show your work and let me check it? It's the Henderson-Hasselbalch equation.

I keep getting about 1.08M for NH4Cl.

I got 9.25527 for the pKa

then for the Henderson-Hasselbalch equation:
9.00=9.25527+log(.600/acid)
10^(-.25527)=(.600/acid)
.5556/.600M=acid
acid=.925925...M
molarity (.925925)*2.10L=

1.94 moles of NH4Cl
1.94 moles *53.4 grams=104.02g of NH4Cl

I think you made an algebra error from line 4 to line 5 in your work.

I got 9.25527 for the pKa
then for the Henderson-Hasselbalch equation:
9.00=9.25527+log(.600/acid)
10^(-.25527)=(.600/acid)correct to this line. Then
0.5556 = 0.6/acid and
acid = 0.6/0.5556 = 1.08 which is the reciprocal of your number)


.5556/.600M=acid
acid=.925925...M
molarity (.925925)*2.10L=

yes you are totally right.

Thank you so much!!!!!!!

Well, I must commend you on your detailed explanation, but unfortunately, your calculations seem to be a little off. Let me help you with the correct answer.

To prepare a buffer solution, we need to use the Henderson-Hasselbalch equation. But before we can do that, we need to find the concentration of NH3 in the solution.

To do that, we can use the equation:

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

Since we want a solution with a pH of 9, that means we have [OH-] = 10^(-5) M.

Kb = (x)(10^(-5)) / (0.600 - x)

Given that Kb = 1.8x10^-5 and solving for x, we can find the concentration of NH3 to be approximately 0.225 M.

Now let's plug this value into the Henderson-Hasselbalch equation:

pH = pKa + log([NH4+]/[NH3])

9.00 = 9.255 + log([NH4+]/0.225)

Solving for [NH4+], we find it to be approximately 0.411 M.

Finally, we can calculate the moles of NH4Cl needed in our solution:

moles = concentration × volume
= 0.411 M × 2.10 L
≈ 0.862 moles

And if we convert this into grams:

mass = moles × molar mass
= 0.862 moles × (14.01 g + 1.01 g + 35.45 g)
≈ 44.8 grams

So, it looks like you made a little error in your calculations. The correct mass of NH4Cl needed to prepare the buffer solution is approximately 44.8 grams. Hope this helps!

To find the correct answer, let's go through the steps together:

1. First, let's calculate the pOH using the given pH of 9.00:
pOH = 14 - pH = 14 - 9.00 = 5.00

2. Convert pOH to OH- concentration using the equation:
pOH = -log[OH-]

=> [OH-] = 10^(-pOH) = 10^(-5.00) = 1.00 x 10^(-5) M

3. Since the solution is a buffer of ammonia (NH3) and ammonium chloride (NH4Cl), we can use the Henderson-Hasselbalch equation to determine the ratio of NH3 to NH4Cl needed to achieve a specific pH. However, we first need to calculate the pKa of NH4Cl.

Using the given Kb for ammonia:
Kb = [NH4+][OH-] / [NH3]
Kb = (x)(x) / (0.600 - x)
Since NH4Cl is a strong electrolyte, we can assume that (0.600 - x) ≈ 0.600
So, Kb ≈ (x)(x) / 0.600
Rearranging the equation:
x^2 = Kb * 0.600
x^2 = (1.8 x 10^(-5)) * 0.600
x^2 = 1.08 x 10^(-5)
x ≈ √(1.08 x 10^(-5))
x ≈ 3.29 x 10^(-3) M

Therefore, the pKa of NH4Cl = -log(Kb) = -log(1.8 x 10^(-5)) ≈ 4.7446

4. Now, we can use the Henderson-Hasselbalch equation with the given pH and pKa to find the ratio of NH3 to NH4Cl:
pH = pKa + log ([NH3] / [NH4+])
9.00 = 4.7446 + log ([NH3] / [NH4+])

Rearranging the equation:
log ([NH3] / [NH4+]) = 9.00 - 4.7446
log ([NH3] / [NH4+]) = 4.2554

Now, we can convert the ratio into actual concentrations using the given total volume of 2.10 L:
[NH3] / [NH4+] = antilog(4.2554)
[NH3] / [NH4+] ≈ 169.94

Since the total concentration of NH3 and NH4+ together is 0.600 M, we can set up an equation to solve for each concentration:
[NH4+] = x
[NH3] = 169.94x

Substituting into the total concentration equation:
x + 169.94x = 0.600
170.94x = 0.600
x ≈ 0.0035 M (concentration of NH4+)
169.94x ≈ 0.595 M (concentration of NH3)

5. Finally, calculate the number of moles of NH4Cl needed:
Moles of NH4Cl = volume of solution (in L) * concentration of NH4+
Moles of NH4Cl = 2.10 L * 0.0035 M ≈ 0.00735 moles

6. Convert moles of NH4Cl to grams using its molar mass:
Molar mass of NH4Cl = 14.01 g/mol (molar mass of NH3) + 1.01 g/mol (molar mass of H) + 35.45 g/mol (molar mass of Cl)
Molar mass of NH4Cl ≈ 53.46 g/mol

Grams of NH4Cl = moles of NH4Cl * molar mass of NH4Cl
Grams of NH4Cl = 0.00735 moles * 53.46 g/mol ≈ 0.392 grams (rounded to three decimal places)

Therefore, the correct answer is approximately 0.392 grams of dry NH4Cl needed to add to 2.10 L of the 0.600 M NH3 solution to prepare a buffer with a pH of 9.00.