a solution containing 100.0 mL of 0.135 M CH3COOH (ka=1.8E-5) is being titrated with 0.54 M NaOH. calculate the ph:
at the equivalence point
To calculate the pH at the equivalence point of the titration, we need to determine the concentration of the resulting solution after the reaction between CH3COOH and NaOH has occurred.
At the equivalence point, the moles of CH3COOH will be equal to the moles of NaOH added. This means that the amount of moles of CH3COOH that have reacted with NaOH will be equal to the amount of moles of NaOH added.
Let's start by calculating the number of moles of CH3COOH in the initial solution:
Moles of CH3COOH = Volume of CH3COOH solution (in L) x Concentration of CH3COOH (in mol/L)
Given:
Volume of CH3COOH solution = 100.0 mL = 0.100 L
Concentration of CH3COOH = 0.135 M
Moles of CH3COOH = 0.100 L x 0.135 mol/L = 0.0135 mol
Since CH3COOH and NaOH react in a 1:1 ratio, the moles of NaOH will also be 0.0135 mol at the equivalence point.
Now, we need to determine the concentration of the resulting solution. This can be done by calculating the total volume of the solution after the reaction and dividing the moles of solute by this volume.
Considering that the volume of NaOH added at the equivalence point is not provided, we cannot calculate the exact volume of the resulting solution. However, we can use the assumption that the volumes of CH3COOH and NaOH are additive.
Let's assume that the volume of NaOH added is V mL. Therefore, the total volume of the resulting solution will be 100.0 mL + V mL.
Using the assumption of additivity, we can calculate the concentration of CH3COOH in the resulting solution:
Concentration of CH3COOH (in mol/L) = Moles of CH3COOH / Total volume of the resulting solution (in L)
Concentration of CH3COOH = 0.0135 mol / (0.100 L + V mL)
Now, since CH3COOH is a weak acid, its reaction with NaOH forms water and a sodium acetate (CH3COONa) salt. At the equivalence point, all of the CH3COOH has reacted, and only the sodium acetate remains in the resulting solution.
The sodium acetate will undergo hydrolysis, resulting in the formation of acetate ions (CH3COO-) and an increase in the pH of the solution.
To find the pH, we need to calculate the concentration of acetate ions (CH3COO-) in the resulting solution. This can be done using the Henderson-Hasselbalch equation:
pH = pKa + log([CH3COO-] / [CH3COOH])
Given:
Ka (acid dissociation constant) of CH3COOH = 1.8E-5 (from the question)
pKa = -log(Ka)
We can rewrite the Henderson-Hasselbalch equation as:
pH = -log(Ka) + log([CH3COO-] / [CH3COOH])
Now, let's substitute the values:
pH = -log(1.8E-5) + log([CH3COO-] / [CH3COOH])
To find the concentration of acetate ions ([CH3COO-]) and CH3COOH in the resulting solution, we need to consider their reaction stoichiometry.
CH3COOH (acetic acid) + NaOH (sodium hydroxide) → CH3COONa (sodium acetate) + H2O (water)
The balanced chemical equation tells us that 1 mole of CH3COOH reacts with 1 mole of NaOH to produce 1 mole of CH3COONa.
Thus, at the equivalence point, the concentration of CH3COOH will be zero (since all of it has reacted).
Now, the concentration of acetate ions ([CH3COO-]) will be equal to the concentration of the sodium acetate (CH3COONa) formed.
Since the moles of NaOH added equal the moles of CH3COOH initially present, the moles of sodium acetate formed will also be equal to the moles of NaOH added.
Moles of sodium acetate = moles of NaOH added = 0.0135 mol
The total volume of the resulting solution is the sum of the initial volume of CH3COOH (100.0 mL = 0.100 L) and the volume of NaOH added (V mL), which can be expressed as (0.100 + V) L.
Now, we can calculate the concentration of the acetate ions:
Concentration of CH3COO- = Moles of sodium acetate / Total volume of the resulting solution
Concentration of CH3COO- = 0.0135 mol / (0.100 L + V mL)
Finally, we substitute the values into the Henderson-Hasselbalch equation to find the pH:
pH = -log(1.8E-5) + log(Concentration of CH3COO- / Concentration of CH3COOH)
pH = -log(1.8E-5) + log(0.0135 mol / (0.100 L + V mL))
This is the mathematical calculation for the pH at the equivalence point. To obtain an actual numerical value, you would need to specify the volume of NaOH added (V mL).