Question
For the titration of 50.00 mL of 0.1000 M ammonia
with 0.1000 M HCl, calculate the pH (a) before the
addition of any HCl solution, (b) after 20.00 mL of the
acid has been added, (c) after half of the NH3 has been
neutralized, and (d) at the equivalence point.
with 0.1000 M HCl, calculate the pH (a) before the
addition of any HCl solution, (b) after 20.00 mL of the
acid has been added, (c) after half of the NH3 has been
neutralized, and (d) at the equivalence point.
Answers
a is pure ammonia.
............NH3 + HOH ==> NH4^+ + OH^-
initial.....0.1M............0.......0
change.......-x..............x.......x
equil......0.1-x.............x........x
Kb = (NH4)(OH^-)/(NH3)
Solve for x - (OH^-) and convert to pH.
b. millimols NH3 = 50.00 x 0.1000 = 5.000
mmols HCl = 20.00 x 0.1000 = 2.000
You have an excess of NH3 with the formation of NH4Cl. This is a buffer. Use the Henderson-Hasselbalch equation.
c. Done the same way as b.
d. At the equivalence point the pH is determined by the hydrolysis of the salt, NH4Cl.
Write the hydrolysis equation, set up an ICE as above, and solve for H3O^+ and pH.
............NH3 + HOH ==> NH4^+ + OH^-
initial.....0.1M............0.......0
change.......-x..............x.......x
equil......0.1-x.............x........x
Kb = (NH4)(OH^-)/(NH3)
Solve for x - (OH^-) and convert to pH.
b. millimols NH3 = 50.00 x 0.1000 = 5.000
mmols HCl = 20.00 x 0.1000 = 2.000
You have an excess of NH3 with the formation of NH4Cl. This is a buffer. Use the Henderson-Hasselbalch equation.
c. Done the same way as b.
d. At the equivalence point the pH is determined by the hydrolysis of the salt, NH4Cl.
Write the hydrolysis equation, set up an ICE as above, and solve for H3O^+ and pH.
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