What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.010 M CH3CO2Na? Assume the volumes of the solution are additive and that Ka = 1.8 x 10-5 for CH3CO2H.
4. How is this solved?
The correct answer is 3.74 if you are wondering
Use the Henderson-Hasselbalch equation and substitute millimols for base and millimols for acid. Solve for pH.
To solve this problem, we will use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka)
[A-] = the concentration of the conjugate base
[HA] = the concentration of the acid
First, let's calculate the concentrations of the acid (CH3CO2H) and the conjugate base (CH3CO2-) after mixing the solutions:
For CH3CO2H:
Concentration = moles / volume
Moles = concentration x volume
[CH3CO2H] = (0.10 M) x (0.02500 L) = 0.0025 mol
For CH3CO2Na:
Concentration = moles / volume
Moles = concentration x volume
[CH3CO2-] = (0.010 M) x (0.02500 L) = 0.00025 mol
Now, let's substitute these values into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
pH = -log(Ka) + log ([CH3CO2-] / [CH3CO2H])
We are given that Ka = 1.8 x 10^-5.
pH = -log(1.8 x 10^-5) + log (0.00025 / 0.0025)
Calculating this expression:
pH = -(-4.74) + log (0.00025 / 0.0025)
pH = 4.74 + log (0.0001)
Using logarithm rules:
pH = 4.74 - 4
pH = 0.74
So, the pH of the solution is 0.74.
To solve this question, we need to understand how the CH3CO2H and CH3CO2Na react in solution.
CH3CO2H is a weak acid, and when it dissolves in water, it releases H+ ions. CH3CO2Na, on the other hand, is a salt that dissociates into CH3CO2- ions and Na+ ions.
When we mix the two solutions, CH3CO2H and CH3CO2Na, they will undergo a reaction known as a hydrolysis reaction. In this reaction, the CH3CO2- ions from the CH3CO2Na solution will react with the H+ ions from the CH3CO2H solution, forming CH3CO2H and water.
Let's calculate the moles of CH3CO2H and CH3CO2Na:
Moles of CH3CO2H = (volume in liters) x (molarity) = (0.025 L) x (0.10 mol/L) = 0.0025 mol
Moles of CH3CO2Na = (volume in liters) x (molarity) = (0.025 L) x (0.010 mol/L) = 0.00025 mol
Since CH3CO2H is a weak acid, it does not dissociate completely. Its dissociation can be represented as follows:
CH3CO2H + H2O ⇌ CH3CO2- + H3O+
The equilibrium expression for the dissociation of CH3CO2H is given by:
Ka = [CH3CO2-] x [H3O+] / [CH3CO2H]
We are given that Ka = 1.8 x 10^-5.
Now, let's calculate the concentration of CH3CO2- and H3O+ ions in the final solution after the reaction:
First, assume that x moles of CH3CO2H react to form x moles of CH3CO2- and H3O+. Since the initial concentration of CH3CO2H is 0.0025 mol and x moles react, the final concentration of CH3CO2H will be (0.0025 - x) mol.
Using the equilibrium expression, we have:
1.8 x 10^-5 = (x)(x) / (0.0025 - x)
Solving this equation will give us the value of x, which represents the concentration of CH3CO2- and H3O+ ions in moles per liter. Finally, we can calculate the pH using the equation:
pH = -log[H3O+]
This is how you would solve this question. However, the calculation involves solving a quadratic equation, so the final answer might require the use of a calculator or software.