Calculate how many milliliters of 0.100 M HCl must be added to 25.0 ml of a 0.200 M solution of NH3 to prepare a buffer that has a pH of 9.50? Kb = 1.76 x 10^-5
.............NH3 + HCl ==> NH4Cl
initial.....5 mmoles..0.....0
add..................x.........
change......-x.......x.......+x
equil........5.x......0.......+x
Plug that into the Hederson-Hasselbalch equation.
pH = pKa + log (base/(acid)
Substitute into the HH equation and solve for x which will be mmols HCl.
Then M = mmoles/mL and solve for mL.
Post your work if you get stuck.
To calculate how many milliliters of 0.100 M HCl must be added to the NH3 solution to prepare a buffer with a specific pH, we need to use the Henderson-Hasselbalch equation for a buffer:
pH = pKa + log ([A-] / [HA])
In this case, NH3 acts as a base (A-) and NH4+ acts as its conjugate acid (HA). The pKa of NH4+ is related to the Kb of NH3 by the equation pKa + pKb = 14.
Given:
pH = 9.50
Kb = 1.76 x 10^-5
Volume of NH3 solution = 25.0 ml
Concentration of NH3 (A-) = 0.200 M
First, calculate the pKa:
pKb = -log(Kb)
pKb = -log(1.76 x 10^-5)
pKb = 4.75
Using the equation pKa + pKb = 14, we can find the pKa:
pKa = 14 - 4.75
pKa = 9.25
Now we can use the Henderson-Hasselbalch equation to find the ratio [A-] / [HA]:
9.50 = 9.25 + log ([A-] / [HA])
Rearrange the equation:
log ([A-] / [HA]) = 9.50 - 9.25
log ([A-] / [HA]) = 0.25
Convert the equation to exponential form:
[A-] / [HA] = 10^0.25
[A-] / [HA] = 1.78
Since the concentration of NH3 (A-) is given as 0.200 M, we can set up the following equation:
0.200 M / [HA] = 1.78
Rearrange and solve for [HA]:
[HA] = 0.200 M / 1.78
[HA] ≈ 0.112 M
Next, we need to find the amount of NH4+ (HA) that will be formed from the NH3 solution:
[HA] = moles / volume
0.112 M = moles / 25.0 ml
Rearrange and solve for moles:
moles = 0.112 M × 25.0 ml
moles ≈ 2.80 mmol
Since the reaction between NH3 and HCl is 1:1, we need the same amount of moles of HCl as NH4+ moles.
Now we can calculate the volume of 0.100 M HCl solution needed using the concentration and stoichiometry:
Volume of HCl solution = moles / concentration
Volume of HCl solution = 2.80 mmol / 0.100 M
Convert the moles to liters (divide by 1000) and rearrange the equation:
Volume of HCl solution = (2.80 mmol / 0.100 M) / 1000
Volume of HCl solution ≈ 0.028 L
Convert the volume to milliliters:
Volume of HCl solution = 0.028 L × 1000
Volume of HCl solution ≈ 28.0 ml
Therefore, approximately 28.0 milliliters of 0.100 M HCl must be added to 25.0 milliliters of the 0.200 M NH3 solution to prepare a buffer with a pH of 9.50.