For the titration of 20.0 mL of 0.1500 mol/L NH3(aq) with 0.1500 mol/L HI(aq)(the titrant), calculate
a) the pH before any HI(aq) is added
b) the pH at the equivalence point
At the beginning of the titration (before any HI is added), you have pure 0.1500 M NH3(aq).
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3(g)
Set up an ICE chart and calculate OH^-, the pOH and pH from that.
At the equivalence point, what do you have.
NH3 + HCl ==> NH4Cl in water.
So set up the hydrolysis of NH4^+, use Ka and solve for the *H^+) and pH.
NH4^+ + HOH ==> NH3 + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4+)
To calculate the pH before any HI(aq) is added, we need to determine the concentration of OH- ions in the NH3 solution.
a) pH before any HI(aq) is added:
For NH3(aq), it acts as a weak base and the reaction can be represented as follows:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
The initial concentration of NH3 is 0.1500 mol/L. In a 1:1 stoichiometric ratio, NH3 reacts with OH- to form NH4+. Therefore, the initial concentration of OH- ions is also 0.1500 mol/L.
Since NH3 is a weak base, we can use the formula for calculating the pOH of a weak base solution:
pOH = -log[OH-]
Substituting the value of [OH-]:
pOH = -log(0.1500) = 0.823
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH = 14 - 0.823 = 13.177
Therefore, the pH before adding any HI(aq) is approximately 13.177.
b) pH at the equivalence point:
For the reaction between NH3 and HI, it can be represented as:
NH3(aq) + HI(aq) ⇌ NH4I(aq)
At the equivalence point, the moles of NH3 and HI are equal, meaning that NH3 has been fully reacted. NH3 is a weak base, while HI is a strong acid. Therefore, the resulting solution is acidic.
Since HI is a strong acid, it dissociates completely, producing H+ ions. The concentration of H+ ions is equal to the initial concentration of HI, which is 0.1500 mol/L.
To calculate the pH at the equivalence point, we can use the formula for calculating the negative logarithm of H+ concentration:
pH = -log[H+]
Substituting the value of [H+]:
pH = -log(0.1500) = 0.823
Therefore, the pH at the equivalence point is approximately 0.823.
To calculate the pH before any HI(aq) is added, we need to consider the dissociation of NH3 in water. NH3 is a weak base which reacts with water to form NH4+ and OH- ions.
a) To find the pH before any HI(aq) is added, we need to determine the concentration of OH- ions in the NH3 solution.
Step 1: Write the equation for the dissociation of NH3 in water:
NH3 + H2O ⇌ NH4+ + OH-
Step 2: Construct the equilibrium expression for the dissociation of NH3:
Kb = [NH4+][OH-] / [NH3]
Step 3: Find the value of Kb for NH3. The Kb value for NH3 is 1.8 x 10^-5 at 25°C.
Step 4: Write the balanced equation for the dissociation of NH4+:
NH4+ ⇌ NH3 + H+
Step 5: Calculate the initial concentration of NH3:
0.1500 mol/L × 0.0200 L = 0.003 mol
Step 6: Assume x mol/L of NH3 reacts with water and produces x mol/L of NH4+ and OH-. Since NH3 is a weak base, we can make the approximation that the concentration of NH4+ and OH- is x mol/L.
Step 7: Calculate the equilibrium concentration of NH3:
[NH3] = 0.003 - x
Step 8: Calculate the equilibrium concentration of NH4+ and OH- using the equation for the dissociation of NH3:
[NH4+] = [OH-] = x
Step 9: Substitute the concentrations into the equilibrium expression Kb and solve for x:
1.8 x 10^-5 = (x)(x) / (0.003 - x)
Solve this quadratic equation to find the value of x.
Step 10: Calculate the concentration of OH- ions using the value of x:
[OH-] = x
Step 11: Calculate the pOH:
pOH = -log[OH-]
Step 12: Calculate the pH using the pOH:
pH = 14 - pOH
To calculate the pH at the equivalence point, we need to consider the reaction between NH3 (base) and HI (acid) when they react in a 1:1 stoichiometric ratio.
b) At the equivalence point, the moles of NH3 reacting with HI are equal in a 1:1 ratio. Therefore, the concentration of NH3 and HI will be the same at the equivalence point.
Since both NH3 and HI react to form NH4+ and I- ions in water, we can see that the concentration of NH4+ will be equal to the concentration of I- at the equivalence point.
Given the initial concentration of NH3 as 0.1500 mol/L, we can determine that the concentration of NH4+ at the equivalence point is also 0.1500 mol/L.
To calculate the pH at the equivalence point, we need to consider the dissociation of NH4+ in water.
Step 1: Write the equation for the dissociation of NH4+ in water:
NH4+ ⇌ NH3 + H+
Step 2: Construct the equilibrium expression for the dissociation of NH4+:
Ka = [NH3][H+] / [NH4+]
Step 3: Find the value of Ka for NH4+. The Ka value for NH4+ is 5.6 x 10^-10 at 25°C.
Step 4: Calculate the equilibrium concentration of NH3 and H+ using the concentration of NH4+ at the equivalence point:
[NH3] = [H+] = 0.1500 mol/L
Step 5: Substitute the concentrations into the equilibrium expression Ka and solve for [H+].
Step 6: Calculate the pH using the concentration of [H+]:
pH = -log[H+]
Follow these steps to get the values for pH before any HI(aq) is added and pH at the equivalence point.