Question

Find the volume (mL) and the pH of 0.135 M HCl needed to reach the equivalence point(s) in titrations of 57.2 mL of 0.226 M NH3. (You need to find the pH at the equivalence point, not the initial pH of the solution.) Kb of NH3 = 1.76 x 10^-5

Answer 1

At the equivalence point moles of HCl added = moles of OH- present.

NH3 + H2O ==> NH4+   + OH-

Moles of NH3 present = 0.0572 L *0.226 M = 0.013 moles

	NH3	H2O	NH4+	OH-
initial	0.013 moles	--	0	0
change	-x	--	+x	+x
equilibrium	0.013-x	--	+x	+x

Kb = [NH4+][OH-]/[NH3]

or, 1.76 x 10^-5 = x*x/(0.013-x)

or, x = 0.0005

[OH-] = 0.0005 moles

moles of HCl required = 1000mL *0.0005/0.135 = 3.7 mL

Volume of HCl required = 3.7 mL.

At equivalence point amount of acid and base is equal . So pH = 7