What is the pH of the solution created by combining 12.00 mL of the 0.10 M base with 8.00 mL of the 0.10 M acid
acid: acetic acid
base: ammonia
This is similiar to the other acid/base problem you posted. You can work this one when you find out how to work that one.
To find the pH of the solution created by combining the given volumes and concentrations of the acid and base, you need to understand the concept of acid-base reactions and their equilibrium. Specifically, we will use the concept of the equilibrium constant, Ka or Kb, to determine the pH.
1. Determine the equation for the reaction:
The reaction between acetic acid (acid) and ammonia (base) can be represented as follows:
CH3COOH (acid) + NH3 (base) ⇌ CH3COO- + NH4+
2. Write the expression for the equilibrium constant (Ka or Kb):
Since acetic acid is a weak acid and ammonia is a weak base, we will use their respective equilibrium constants, Ka and Kb.
For the acid, acetic acid, the expression for equilibrium constant Ka can be written as:
Ka = [CH3COO-][H3O+] / [CH3COOH]
For the base, ammonia, the expression for equilibrium constant Kb can be written as:
Kb = [NH4+][OH-] / [NH3]
3. Determine the values of the equilibrium constants (Ka or Kb):
The values of Ka or Kb can be found in reference books or online sources. For acetic acid, Ka is approximately 1.8 x 10^-5. For ammonia, Kb is approximately 1.8 x 10^-5 (which is equal to Kw / Ka, where Kw is the autoionization constant of water).
4. Calculate the concentration of the products and reactants:
To determine the concentration of the products and reactants, we need to use the given volumes and concentrations.
a. For the acid, acetic acid:
moles of acetic acid = volume (in L) × concentration (in mol/L)
moles of acetic acid = 0.012 L × 0.10 mol/L = 0.0012 mol
b. For the base, ammonia:
moles of ammonia = volume (in L) × concentration (in mol/L)
moles of ammonia = 0.008 L × 0.10 mol/L = 0.0008 mol
5. Calculate the concentration of the products and reactants at equilibrium:
Since the given volumes are small compared to the total volume of the solution, we assume there is no significant change in the total volume. Therefore, we can assume that the total moles of the solute remain the same.
a. For the acid, acetic acid:
Concentration of acetic acid at equilibrium = (initial concentration - moles of acetic acid reacted) / total volume
Concentration of acetic acid at equilibrium = (0.10 mol/L - 0.0012 mol) / (0.012 L + 0.008 L) = 0.099 mol/L
b. For the base, ammonia:
Concentration of ammonia at equilibrium = (initial concentration - moles of ammonia reacted) / total volume
Concentration of ammonia at equilibrium = (0.10 mol/L - 0.0008 mol) / (0.012 L + 0.008 L) = 0.099 mol/L
6. Plug the equilibrium concentrations into the equilibrium constant expression:
Since the concentration of H3O+ (from acetic acid) and OH- (from ammonia) are not given, we can assume their concentrations are initially very small, leading to negligible change during the reaction. Therefore, we can assume their concentrations at equilibrium are zero.
For the acid, acetic acid:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Ka = ([CH3COO-] × 0) / [CH3COOH]
Ka = 0
For the base, ammonia:
Kb = [NH4+][OH-] / [NH3]
Kb = ([NH4+] × 0) / [NH3]
Kb = 0
7. Determine the pH:
Since the value of Ka or Kb is zero, the concentration of H3O+ and OH- ions is very low, so the pH can be calculated using the equation:
pH = -log[H3O+]
In this case, since the concentration of H3O+ is zero, the pH is undefined.
Therefore, the pH of the solution created by combining 12.00 mL of the 0.10 M acetic acid and 8.00 mL of the 0.10 M ammonia is undefined.