Consider the titration of 29.00 mL of aqueous ammonia with 18.70 mL of 0.1250 M HCl (aq). Kb for NH3 (aq) is 1.8 x 10^-5.
A.) What is the formula of the solute at the equivalence point?
B.) What is the balanced equation for the ionization of the cation of the solute?
C.) What is the molarity of the solute at the equivalence point?
D.) What is the pH at the equivalence point?
You must recognize what is going on in the titration. The equation is
NH3.H2O + HCl ==> NH4Cl + H2O
Obviously NH4Cl is what you have at the equivalence point.
Hydrolyze NH4Cl to determine the pH of th solute.
NH4^+ + H2O ==> H3O^+ + NH3
Set up the ICE chart, and plug all into Ka for the hydrolysis. Ka = (Kw/Kb) =
(NH3)(H3O^+)/(NH4^+)
molarity of solute is moles/L.
Post your work if you get stuck.
I understand the ICE chart, but I'm unsure where you get the values to put in the chart?
This is hydrolysis of 'salt' ions... After titration to equivalence point, only salt remains in solution. Before hydrolysis, only salt ions will be present. Using NH4Cl => NH4+ + Cl- only the NH4+ will under go hydrolysis because theoretically Cl- + HOH => HCL + OH-. However, HCl being a strong acid (electrolyte) prefers to remain 100% ionized. This leaves:
NH4+ + HOH => NH4OH + H+
Ci: [salt] --- 0 0
∆C: -x --- +x +x
Ceq [salt-x]* --- x x
~[salt]
----
*you can drop 'x' under NH4+
----
Apply equilibrium row to Ka expression for Hydrolysis of NH4+.
Ka(NH4+) = (Kw/Kb)
=([NH4OH][H+])/([NH4+)]
=(x)(x)/[NH4+] = x^2/[NH4+]
x = [H+] = ((Kw/Kb)[NH4+])^1/2
pH at equivalence point = -log[H+]
To answer these questions, we need to analyze the given information and apply the concepts of titration and equilibrium.
A.) To determine the formula of the solute at the equivalence point, we need to understand the reaction that occurs during the titration. In this case, we are titrating aqueous ammonia (NH3) with hydrochloric acid (HCl). The balanced equation for this reaction is:
NH3 (aq) + HCl (aq) → NH4+ (aq) + Cl- (aq)
At the equivalence point, all the ammonia will react fully with the hydrochloric acid, and the concentration of the solute will be the same as the concentration of the NH4+ ion. Therefore, the formula of the solute at the equivalence point is NH4+.
B.) The balanced equation for the ionization of the NH4+ cation can be written as:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
In this equation, NH4+ dissociates to form NH3 (ammonia) and H+ (hydrogen ion). This reaction represents the ionization of the ammonium ion.
C.) To determine the molarity of the solute at the equivalence point, we need to use the concept of stoichiometry and the given volumes and concentrations of the reactants.
First, we need to find the number of moles of HCl used in the titration:
moles of HCl = volume of HCl (L) × concentration of HCl (M)
= 0.0187 L × 0.1250 M
= 0.0023375 mol
Since the balanced equation shows a 1:1 stoichiometric ratio between HCl and NH3, the number of moles of NH3 will also be 0.0023375 mol.
Next, we can calculate the molarity of NH3 at the equivalence point using the volume of NH3 and the number of moles:
molarity of NH3 = moles of NH3 / volume of NH3 (L)
= 0.0023375 mol / 0.0290 L
= 0.0806 M
Therefore, the molarity of the solute (NH3) at the equivalence point is 0.0806 M.
D.) To find the pH at the equivalence point, we need to consider the ionization constant of NH3 (Kb) and the resulting concentration of H+ ions.
The ionization constant of NH3 (Kb) is given as 1.8 x 10^-5. Since Kb is related to the equilibrium constant expression for the dissociation of NH3 as follows:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
We can write the Kb expression as:
Kb = [NH4+][OH-] / [NH3]
At the equivalence point, all the NH3 reacts with HCl, resulting in the formation of NH4+ ions and leaving no NH3 remaining. Therefore, the concentration of NH3 will be zero, and the Kb expression simplifies to:
Kb = [NH4+][OH-] / 0
Since Kb and the equilibrium expression are related through the following equation:
Kw = Kb × Ka
where Kw is the autoionization constant of water (1.0 x 10^-14) and Ka is the ionization constant for water (1.0 x 10^-14), we can rearrange the equation to solve for OH- concentration:
OH- = Kw / Kb
OH- = (1.0 x 10^-14) / (1.8 x 10^-5)
OH- = 5.56 x 10^-10 M
From the equation for the ionization of NH4+:
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
We know that the concentration of H+ and OH- ions are equal in a neutral solution. Therefore, at the equivalence point, the concentration of H+ ions will also be 5.56 x 10^-10 M.
To find the pH, we use the equation:
pH = -log[H+]
pH = -log(5.56 x 10^-10)
pH ≈ 9.26
Therefore, the pH at the equivalence point is approximately 9.26.