a) Find the pH at the equivalence point when 40ml of 0.025 M of methylamine (NH2CH3 and Kb = 4.4x10^-4) is titrated with 0.025 M HCl.
b) The Kf for the complex ion Ag(NH3)2+ is 1.7x10^7. The Ksp for AgI is 1.5x10^-16. What is the molar solubility of AgI in a solution that is 2.0 M in NH3?
c) Find the percent ionization of a 1.00 M solution of formic acid (HCOOH) when Ka = 1.7x10^-4.
a) To find the pH at the equivalence point, we need to determine the moles of methylamine and HCl involved in the reaction.
First, let's calculate the moles of methylamine:
Moles of methylamine = 0.040 L × 0.025 M = 0.0010 mol
Since the balanced equation for the reaction between methylamine (base) and HCl (acid) is:
NH2CH3 + HCl -> NH3CH3+Cl-
We can see that one mole of methylamine reacts with one mole of HCl. Therefore, the moles of HCl will also be 0.0010 mol.
Now, let's calculate the concentration of the NH3CH3+ ion at the equivalence point:
NH3CH3+ is the conjugate acid of the weak base methylamine. To find its concentration, we need to consider the Kb value of methylamine.
Kb = [NH3CH3+][OH-] / [NH2CH3]
Since this is a neutralization reaction, at the equivalence point, [OH-] = [H+].
Therefore, Kb = [NH3CH3+][H+]/[NH2CH3]
Letting x be the concentration of NH3CH3+ at the equivalence point, x[H+] = x^2.
Now, we can solve the quadratic equation:
Kb = x^2 / (0.0010 - x)
Substituting the Kb value and solving for x, we get x = 0.040.
At the equivalence point, the concentration of H+ is equal to the concentration of OH-, which can be used to find the pOH:
pOH = -log10(0.040) ≈ 1.40
Finally, since pH + pOH = 14, we can find the pH at the equivalence point:
pH = 14 - 1.40 ≈ 12.60
Therefore, the pH at the equivalence point is approximately 12.60.
b) To find the molar solubility of AgI in a solution that is 2.0 M in NH3, we need to calculate the concentration of Ag+ ions.
The formation of Ag(NH3)2+ from AgI can be represented by the following balanced equation:
AgI + 2NH3 -> Ag(NH3)2+ + I-
The Kf value for the complex ion Ag(NH3)2+ will help us determine the concentration of Ag+ ions.
Kf = [Ag(NH3)2+]/([Ag+][NH3]^2)
Substituting the given Kf value, we have:
1.7x10^7 = [Ag(NH3)2+]/[(Ag+)(2.0 M)^2]
To find [Ag(NH3)2+], we need to find the concentration of Ag+ at equilibrium.
Since AgI is a sparingly soluble compound, we can assume that the concentration of Ag+ at equilibrium will be much lower than the initial concentration of AgI, and therefore, we can neglect the change in Ag+ concentration from the NH3 complex formation.
Let's assume the concentration of Ag+ at equilibrium is x M. Then, the concentration of Ag(NH3)2+ at equilibrium will also be x M.
Substituting these values, we have:
1.7x10^7 = x / [(x)(2.0 M)^2]
Now, solve this equation for x:
1.7x10^7 = 1 / [(2.0 M)^2]
x = 1 / (1.7x10^7 × (2.0 M)^2)
x ≈ 1.7x10^-8 M
Therefore, the molar solubility of AgI in a solution that is 2.0 M in NH3 is approximately 1.7x10^-8 M.
c) To find the percent ionization of a 1.00 M solution of formic acid (HCOOH) when Ka = 1.7x10^-4, we need to determine the concentration of H+ ions produced and calculate the percent ionization.
The balanced equation for the dissociation of formic acid is:
HCOOH -> H+ + HCOO-
Let's assume that x M of formic acid dissociates into H+ ions.
Using the Ka expression for formic acid, we have:
Ka = [H+][HCOO-]/[HCOOH]
Since we are assuming x M of HCOOH dissociates, the concentration of HCOOH remaining will be (1.00 - x) M.
Therefore, Ka = x * x / (1.00 - x)
Now, solve this quadratic equation for x:
1.7x10^-4 = x * x / (1.00 - x)
Simplifying, we get:
1.7x10^-4 (1.00 - x) = x^2
Rearranging, we have:
1.7x10^-4 - 1.7x10^-4x = x^2
This equation can be solved using methods like factoring or quadratic formula. Solving it, we find that x ≈ 0.0097 M.
The percent ionization can be calculated as:
Percent Ionization = (concentration of H+ ions produced / initial concentration of HCOOH) * 100
Percent Ionization = (0.0097 M / 1.00 M) * 100 ≈ 0.97%
Therefore, the percent ionization of a 1.00 M solution of formic acid when Ka = 1.7x10^-4 is approximately 0.97%.