Asked by Taylor M
Consider the titration of 30.0 mL sample
of 0.050 M NH3 with 0.025 M HCl.
Calculate the pH after the following volumes
of titrant (acid)have been added:
a)0 mL,
b)20.0mL,
c)60.0mL,
d)65mL
Kb for NH3 is 1.8 x 10^-5
of 0.050 M NH3 with 0.025 M HCl.
Calculate the pH after the following volumes
of titrant (acid)have been added:
a)0 mL,
b)20.0mL,
c)60.0mL,
d)65mL
Kb for NH3 is 1.8 x 10^-5
Answers
Answered by
DrBob222
The secret to these is to recognize what you have in the solution at each addition.
The first thing you do is to determine where the equivalence point is; i.e., number of mL HCl added to get to the equivalence point.
20 mL--you have a mixture of NH3 and NH4Cl. That makes a buffer; use the Henderson-Hasselbalch equation.
60 mL is the eq point. The pH is determined by the hydrolysis of the salt NH4Cl.
0 mL you have 0.50M NH3. I'm sure you've worked many a problem like that.
The first thing you do is to determine where the equivalence point is; i.e., number of mL HCl added to get to the equivalence point.
20 mL--you have a mixture of NH3 and NH4Cl. That makes a buffer; use the Henderson-Hasselbalch equation.
60 mL is the eq point. The pH is determined by the hydrolysis of the salt NH4Cl.
0 mL you have 0.50M NH3. I'm sure you've worked many a problem like that.
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