Question

Suppose you titrate 31.1 mL of 2.22 M ammonia, NH3, with 1.55 M HCl.

a) How many moles of NH3 are present in the original NH3 solution?

b)what is the initial pH?

c) what volume of HCl solution is required to reach the equivalence point?

d) what is the pH at the equivalence point?

Answer 1

a)

moles of NH3 = M(NH3)*V(NH3)

= 2.22 M * 0.0311 L

= 0.0690 mol

Answer: 0.0690 mol

b)

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

2.22 0 0

2.22-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*2.22) = 6.321*10^-3

since c is much greater than x, our assumption is correct

so, x = 6.321*10^-3 M

So, [OH-] = x = 6.321*10^-3 M

use:

pOH = -log [OH-]

= -log (6.321*10^-3)

= 2.1992

use:

PH = 14 - pOH

= 14 - 2.1992

= 11.8008

Answer: 11.80

c)

find the volume of HCl used to reach equivalence point

M(NH3)*V(NH3) =M(HCl)*V(HCl)

2.22 M *31.1 mL = 1.55M *V(HCl)

V(HCl) = 44.5432 mL

Answer: 44.5 mL

d)

Given:

M(HCl) = 1.55 M

V(HCl) = 44.5432 mL

M(NH3) = 2.22 M

V(NH3) = 31.1 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 1.55 M * 44.5432 mL = 69.042 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 2.22 M * 31.1 mL = 69.042 mmol

We have:

mol(HCl) = 69.042 mmol

mol(NH3) = 69.042 mmol

69.042 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 69.042 mmol

Volume of Solution = 44.5432 + 31.1 = 75.6432 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 69.042 mmol/75.6432 mL = 0.9127 M

NH4+ + H2O -----> NH3 + H+

0.9127 0 0

0.9127-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.9127) = 2.252*10^-5

since c is much greater than x, our assumption is correct

so, x = 2.252*10^-5 M

[H+] = x = 2.252*10^-5 M

use:

pH = -log [H+]

= -log (2.252*10^-5)

= 4.6475

Answer: 4.65