How many moles of NH4Cl must be added to 1.5L of 0.2 M solution of NH3 to form a buffer whose PH is 9.(Kb=1.8x10^-5)
Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base/(acid)]
You know pH, you can calculate pKa (convert Kb to pKb then pKa + pKb = pKw = 14), you know (NH3), solve for (acid; i.e., NH4Cl). Then M NH4Cl = mols/L. You know M and L, solve for mols. Then mols = grams/molar mass. You know mols and molar mass, solve for grams.
Post your work if you get stuck.
0.5r
0.54
0.36
0.54
To calculate the number of moles of NH4Cl needed to form a buffer with a pH of 9, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log(A-/HA)
In this case, NH3 acts as base (A-) and NH4+ acts as the conjugate acid (HA). So, the equation becomes:
9 = pKa + log(NH3/NH4+)
The pKa of NH3 can be calculated from the Kb value using the equation:
Ka = Kw/Kb
Since we know that Kw = 1.0 x 10^-14 and given that Kb = 1.8 x 10^-5, we can calculate Ka:
Ka = Kw/Kb = (1.0 x 10^-14)/(1.8 x 10^-5) = 5.6 x 10^-10
Now, we can calculate the pKa:
pKa = -log(Ka) = -log(5.6 x 10^-10) = 9.25
Now, we can substitute the pKa value into the Henderson-Hasselbalch equation:
9 = 9.25 + log(NH3/NH4+)
Rearrange the equation:
log(NH3/NH4+) = 9 - 9.25 = -0.25
Take the antilog of both sides:
(NH3/NH4+) = 10^(-0.25)
(NH3/NH4+) = 0.5623
Since the ratio of NH3 to NH4+ in a buffer system is preferably about 1, we can assume that the initial concentrations of NH3 and NH4+ are the same.
Let's assume x moles of NH4Cl are added to the 1.5 L of 0.2 M NH3 solution. So, the final concentration of NH4+ will be x moles in 1.5 L, and the final concentration of NH3 will also be x moles in 1.5 L.
(NH4+ concentration)/(NH3 concentration) = x/(1.5 - x)
Since the ratio of NH3 to NH4+ is 0.5623,
x/(1.5 - x) = 0.5623
Solve this equation to find the value of x:
x = 0.5623(1.5 - x)
x = 0.8435 - 0.5623x
1.5623x = 0.8435
x = 0.539 moles
Therefore, approximately 0.539 moles of NH4Cl must be added to the 1.5L of 0.2 M NH3 solution to form a buffer with a pH of 9.