Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added. (The Kb for NH3 is 1.8e-05.)

Question

(a) 0 mL
.

(b) 25.0 mL
.

(c) 59.0 mL
.

(d) 60.0 mL

(e) 63.0 mL

(f) 67.5 mL

Answer 1

NH3 + HCl ==> NH4Cl
First, calculate the volume HCl needed for the equivalence point. Then divide the "curve" into these portions. You need to where you are on the curve to know how to calculate the pH.
a. beginning point. This is just a weak base, NH3. You know Kb and (NH3), use the
NH3 + HOH ==> NH4^+ + OH^-
Set up an ICE chart and solve for x = OH^- and convert to pH.

b. ALL point between the zero mL and the equivalence point. Use the Henderson-Hasselbalch equation.

c. The equivalence point, Set up an ICE chart for the salt produced at the equivalence, hydrolyze the NH4^+ and use Ka NH4^+ (rememberr that's Kw/Kb for NH3)

d. For all points after the equivalence point use the concn of the excess HCl and convert to pH.