A volume of 25.0 ml of 0.100 M CH3C02H is titrated with 0.100 M NaOH. What is
pH after the addition of 12.5 ml of NaOH? (Ka for CH3CO2H = 1.8x10-5)
Use the Henderson-Hasselbalch equation. However, you will note that the acid is exactly half neutralized; therefore, the base and acid terms will cancel to leave 1, the log of 1 is zero and pH = pKa for an answer but you work it out to see.
To find the pH after the addition of 12.5 ml NaOH, we need to calculate the moles of CH3CO2H and NaOH, determine the limiting reactant, and then find the concentration of the remaining CH3CO2H. From there, we can use the Ka value to find the pH.
Step 1: Calculate the moles of CH3CO2H and NaOH
Moles of CH3CO2H = volume (in L) x concentration = 25.0 ml x (1 L/1000 ml) x 0.100 M = 0.00250 mol
Moles of NaOH = volume (in L) x concentration = 12.5 ml x (1 L/1000 ml) x 0.100 M = 0.00125 mol
Step 2: Determine the limiting reactant
Since the stoichiometric ratio between CH3CO2H and NaOH is 1:1, the limiting reactant is the one with the smallest number of moles. In this case, NaOH has fewer moles, so it is the limiting reactant.
Step 3: Find the moles of remaining CH3CO2H
Moles of remaining CH3CO2H = initial moles - moles reacted with NaOH = 0.00250 mol - 0.00125 mol = 0.00125 mol
Step 4: Calculate the concentration of remaining CH3CO2H
Volume of remaining CH3CO2H = total volume - volume of NaOH added = 25.0 ml - 12.5 ml = 12.5 ml
Concentration of remaining CH3CO2H = moles / volume = 0.00125 mol / (12.5 ml x 1 L/1000 ml) = 0.100 M
Step 5: Calculate the pH using the Ka value
pH = -log[H+]
To find [H+], we need to find [CH3CO2-].
Using the equation for the ionization of CH3CO2H: CH3CO2H ⇌ CH3CO2- + H+
Ka = [CH3CO2-][H+] / [CH3CO2H]
1.8x10^-5 = [CH3CO2-][H+] / 0.100
[H+] = Ka x [CH3CO2H] / [CH3CO2-] = (1.8x10^-5) x (0.100 M) / (0.100 M) = 1.8x10^-5 M
pH = -log(1.8x10^-5) ≈ 4.74
Therefore, the pH after the addition of 12.5 ml of NaOH is approximately 4.74.
To find the pH after the addition of NaOH, we need to determine the concentration of CH3CO2H and CH3CO2-. We can then use the equilibrium expression for the dissociation of CH3CO2H to calculate the concentration of H+ ions, and finally convert it to pH.
First, let's calculate the moles of CH3CO2H initially present in the 25.0 ml of 0.100 M solution. The formula for moles is Moles = Concentration x Volume.
Moles of CH3CO2H = 0.100 M x 0.025 L = 0.0025 moles
Since CH3CO2H dissociates in water to produce H+ and CH3CO2-, the concentration of CH3CO2- is the same as the initial concentration of CH3CO2H.
Now, let's calculate the moles of NaOH added. The formula for moles is Moles = Concentration x Volume.
Moles of NaOH = 0.100 M x 0.0125 L = 0.00125 moles
Since NaOH is a strong base, it will completely dissociate in water to produce Na+ and OH- ions. Therefore, the concentration of OH- ions is the same as the concentration of NaOH.
Knowing the moles of CH3CO2H and CH3CO2-, we can set up the equilibrium expression for the dissociation of CH3CO2H:
CH3CO2H ⇌ CH3CO2- + H+
The equilibrium expression is: Ka = [CH3CO2-][H+]/[CH3CO2H]
Since the initial concentration of CH3CO2H and CH3CO2- is the same (0.0025 moles), we can substitute this value:
Ka = (0.0025 + x)(x) / (0.0025 - x)
where x represents the moles of H+ ions that have formed.
Given that the Ka value for CH3CO2H is 1.8x10-5, we can rearrange the equation:
(1.8x10-5) = (0.0025 + x)(x) / (0.0025 - x)
Solving this quadratic equation will give us the value of x, which represents the moles of H+ ions. We can then convert the concentration of H+ ions to pH using the equation pH = -log[H+].
To summarize, to find the pH after the addition of NaOH, we need to:
1. Calculate the initial moles of CH3CO2H and CH3CO2-.
2. Calculate the moles of NaOH added.
3. Set up the equilibrium expression for CH3CO2H and solve for x.
4. Convert the concentration of H+ ions to pH using the pH equation.