Question

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.00 mL sample of the cleaner, the equivalence point is reached after 11.53 mL of 0.118 M HCL has been added. The initial concentration of ammonia is .108 M. What is the pH of the solution at the equivalence point?

Answer 1

for ammonia Kb = 1.77 * 10^-5.

pOH = pKb + log([NH₄⁺ ] / [NH3])

Calculate the pKb by

pKb = −log(Kb) = - log(1.77 * 10^-5) = 4.75

we are going to calculate pH of solution before adding HCl

Since ammonia is a weak base, you can expect the pH of the solution to be basic, or greater than 7. An equilibrium will be established that will allow you to calculate the concentration of the hydroxide ion, OH−

NH3(aq) + H2O(l) <----> NH₄⁺  (aq) + OH- (aq)
I 0.108 0 0
C (-x) (+x) (+x)

E 0.108-x x x

By definition, the acid dissociation constant will be

Kb = [NH₄⁺][OH−] / [NH3] = (x*x) / (0.108 - x) = x² / (0.108 - x) = 1.77 * 10^-5

x = 1.37 * 10^-3.

Calculate the solution's pOH by

pOH = -log([OH-]) = -log(1.37 * 10^-3) = 2.86

As a result, the pH will be

pH = 14 - pOH = 14 - 2.86 = 11.14

After the addition of 11.53 mL of HCl

Now you start adding acid to the solution. Ammonia will react with the strong acid to produce ammonium chloride, which will act as the conjugate acid → a buffer is formed.

NH3(aq) + HCl(aq) -----> NH4Cl (aq)

The net ionic equation is

NH3 (aq) + H+ (aq) ----->   NH₄⁺ (aq)

number of moles of ammonia initially in solution

Given Volume V1 = 25 ml = 0.025 L M1 = 0.108 mol/L

n1 = 0.108 mol/L * 0.025 L = 0.0027 moles

number of moles of hydrochloric acid added

Given Volume V2 = 11.53 ml = 0.01153 L M2 = 0.118 mol/L

n1 = 0.118 mol/L * 0.01153 L = 0.00136 moles

All HCl will be consumed in the reaction, which means that the number of moles of ammonia will decrease by 0.00136, and 0.00136 moles of NH₄⁺ will be produced.

The volume of the solution will now be

Vsol = 25 + 11.53 =36.53 mL = 0.03653 L

The concentrations of ammonia and ammonium chloride will be

[NH3]=(0.0027 - 0.00136) moles / 0.03653 L = 0.03668 M

[NH₄⁺] = 0.00136 moles / 0.03653 L = 0.03723 M

The pOH of the solution will be

pOH = pKb + log([NH₄⁺]/[NH3]) = 4.75 + log ( 0.03723 M / 0.03668 M )

pOH = 4.756

pH = 14 - 4.7565 = 9.2435