Question

Ammonia (NH₃) is a popular window cleaner. It is also a base.

A. What is the concentration of ammonia if 15.34 mL of 1.26 M HCl is needed to titrate a 10.00 mL sample of the cleaner?

B. Suppose that the sample was diluted to about 50 mL with deionized water prior to the titration to make it easier to mount a pH electrode in it. What effect did this dilution have on the volume of titrant needed? (multiple choice)

- More titrant is needed to reach the equilibrium point?

- Less titrant is needed to reach the equilibrium point?

or

- No change in the amount of titrant is needed to change the equilibrum point?

Answer 1

A)

consider the reaction

HCl + NH3 --> NH4Cl

we know that

moles of HCl added = moles of NH3 present

now

moles = molarity x volume

so

MHcl x VHcl = MNh3 x VNH3

1.26 x 15.34 = MNh3 x 10

MNh3 = 1.93284 M

so

the concentration of ammonia is 1.93284 M

B)

now, given that

the sample of ammonia is diluted

we know dilution decreases the concentration

but the volume increases too

as a result

MNH3 x VNH3 remains constant

so

MHCl x VHCl = MNH3 x VNh3 = constant

so

no change in the amount of titrant is needed to change the equilibrium point