How many grams of NH4Cl must be added to 0.490 L of 1.70 M NH3 solution to yield a buffer solution with a pH of 9.40? Assume no volume change. Kb for NH3 = 1.8 10-5.
pH = pKa + log[(base)/(acid)]
Convert Kb for NH3 into Ka and pKa, substitute into the H-H equation (1.70 for (base), and solve for (acid) in moles/L. Then convert to grams NH4Cl in 0.490 L. Post your work if you get stuck.
I still have it wrong. I have pKa = 9.25. Solved for acid and got 2.39. Then I converted to grams and 62.25. Here's my work ka = 1.0e-14/1.8e-5 = 5.56e-10. solved for pKa -log(5.56e-10) = 9.25. H-H = 9.40=9.25 + log(1.70/acid). acid = 2.39. then, 2.39 x 0.490L = 1.17 M. then 1.17 x 53.4g = 62.65g
To find the answer to this question, we need to calculate the amount of NH4Cl required to create a buffer solution with a pH of 9.40.
First, let's determine the concentration of NH4Cl needed in the buffer solution. Since we want to create a buffer with a pH of 9.40, which is basic, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, NH3 acts as the base (A-) and NH4+ acts as the acid (HA). The pKa can be calculated using the Kb value for NH3:
Kb = Kw / Ka
Since Kw = 1.0 x 10^-14, we can calculate Ka:
Ka = Kw / Kb
Now, we can calculate the pKa:
pKa = -log(Ka)
Using the given Kb value of 1.8 x 10^-5, we find:
Ka = (1.0 x 10^-14) / (1.8 x 10^-5) = 5.6 x 10^-10
pKa = -log(5.6 x 10^-10) ≈ 9.25
Next, we can substitute the values into the Henderson-Hasselbalch equation:
9.40 = 9.25 + log([NH3]/[NH4+])
Since we know that [NH3] = 1.70 M and [NH4+] is the concentration of NH4Cl, we can rearrange the equation to solve for [NH4+]:
log([NH4+]) = 9.40 - 9.25
log([NH4+]) = 0.15
Now, we can convert the log value to an actual concentration. Since the pH scale is logarithmic, we can use the antilog or exponential function to find the concentration:
[NH4+] = 10^(0.15)
[NH4+] ≈ 1.41 M
So, the concentration of NH4Cl needed in the buffer solution is approximately 1.41 M.
Finally, to find the number of grams of NH4Cl needed, we can use the formula:
mass (g) = volume (L) x concentration (M) x molar mass (g/mol)
The volume is given as 0.490 L, and we now know the concentration is 1.41 M. The molar mass of NH4Cl is approximately 53.49 g/mol.
mass (g) = 0.490 L x 1.41 M x 53.49 g/mol
mass (g) ≈ 36.30 g
Therefore, approximately 36.30 grams of NH4Cl must be added to 0.490 L of the 1.70 M NH3 solution to yield a buffer solution with a pH of 9.40.