This is a buffer problem.
Use the Henderson-Hasselbalch equation.
calculate the pH in a solution prepared by dissolving .10 mol of solid NH4Cl in .500 L of .40 M NH3. Assuming no volume change.
3 answers
calculate the pH in a solution prepared by dissolving .10 mol of solid NH4Cl in .500 L of .40 M NH3. Assuming no volume change. Without the use of Henderson-hasselbalch
The HH equation is much easier but here is the other.
(NH3) = 0.40M
(NH4Cl) = (NH4^+) = (Cl^-) = mols/L = 0.1mol/0.500L = 0.2M
...........NH3 + HOH ==> NH4^+ + OH^-
I........0.40..........0.20.......0
C..........-x.............x.......x
E.........0.40-x.......9,.20+x....x
Kb = (NH4^+)(OH^-)/(NH3)
Substitute the E line into Kb above and solve for x = (OH^-). Convert that to pH.
Note: The NH4Cl is a common ion to the NH3 solution (the NH4^+ is the common ion) and the addition of the NH4^+ forces the ionization due to Kb to the left and that limits the basicity. You can assume, to make the solution easier, that 0.200+x = 0.200 and 0.4-x = 0.4.
(NH3) = 0.40M
(NH4Cl) = (NH4^+) = (Cl^-) = mols/L = 0.1mol/0.500L = 0.2M
...........NH3 + HOH ==> NH4^+ + OH^-
I........0.40..........0.20.......0
C..........-x.............x.......x
E.........0.40-x.......9,.20+x....x
Kb = (NH4^+)(OH^-)/(NH3)
Substitute the E line into Kb above and solve for x = (OH^-). Convert that to pH.
Note: The NH4Cl is a common ion to the NH3 solution (the NH4^+ is the common ion) and the addition of the NH4^+ forces the ionization due to Kb to the left and that limits the basicity. You can assume, to make the solution easier, that 0.200+x = 0.200 and 0.4-x = 0.4.
4 answers