Calculate the pH at the equivalence point of 25.0 mL of a 0.100 M acetic acid, CH3COOH, titrated with a 0.100 M sodium hydroxide, NaOH. Ka of acetic acid is 1x10^-5
To calculate the pH at the equivalence point, we need to determine the concentration of the acetate ion, CH3COO-, which is the conjugate base of the acetic acid.
At the equivalence point, the molar amount of moles of CH3COOH will be equal to the moles of NaOH. Since the molar ratio between them is 1:1, we can calculate the moles of NaOH and then determine the concentration of CH3COO-.
Given:
Volume of CH3COOH = 25.0 mL = 0.0250 L
Molarity of CH3COOH = 0.100 M
Molarity of NaOH = 0.100 M
Step 1: Calculate moles of CH3COOH.
moles of CH3COOH = volume (L) x molarity
moles of CH3COOH = 0.0250 L x 0.100 M
moles of CH3COOH = 0.0025 mol
Step 2: Calculate moles of NaOH.
Since the stoichiometric ratio between CH3COOH and NaOH is 1:1, the moles of NaOH will also be 0.0025 mol.
Step 3: Calculate the concentration of CH3COO-.
Since the initial volume of CH3COOH is 25.0 mL and it reacts with an equal volume of NaOH, the total volume at the equivalence point will be 50 mL, or 0.0500 L.
moles of CH3COO- = moles of NaOH = 0.0025 mol
concentration of CH3COO- = moles / volume
concentration of CH3COO- = 0.0025 mol / 0.0500 L
concentration of CH3COO- = 0.050 M
Step 4: Calculate pOH at the equivalence point.
pOH = -log10(concentration of OH-)
Since NaOH is a strong base, it dissociates completely into OH- ions. Therefore, the concentration of OH- is equal to the concentration of NaOH.
pOH = -log10(0.100 M)
pOH = 1
Step 5: Calculate pH at the equivalence point.
Since pH + pOH = 14, we can subtract the pOH from 14 to find the pH.
pH = 14 - pOH
pH = 14 - 1
pH = 13
Therefore, the pH at the equivalence point of the titration is 13.
To calculate the pH at the equivalence point of the titration, we first need to determine the number of moles of acetic acid and sodium hydroxide that react at the equivalence point.
We have 25.0 mL of a 0.100 M acetic acid solution. To find the number of moles, we use the equation:
moles = concentration × volume
moles of acetic acid = 0.100 M × 25.0 mL = 0.100 mol
Since acetic acid is a monoprotic acid, it reacts in a 1:1 ratio with sodium hydroxide. Therefore, at the equivalence point, the number of moles of sodium hydroxide will also be 0.100 mol.
Next, we need to determine the composition of the resulting solution after the reaction has occurred. At the equivalence point, the acetic acid will react completely with the sodium hydroxide to form sodium acetate, NaCH3COO, and water, H2O.
CH3COOH + NaOH → NaCH3COO + H2O
The sodium acetate will dissociate in water to form sodium ions and acetate ions. Since sodium acetate is a salt of a weak acid (acetic acid), its acetate ion will hydrolyze and react with water to form hydroxide ions. This hydrolysis reaction contributes to the basic nature of the solution. The NaOH added during the titration also contributes to the basicity of the solution.
Due to this hydrolysis reaction, we have a basic solution, and the pH is calculated using the equation:
pOH = -log10[OH-]
To determine the pOH, we need to find the concentration of hydroxide ions ([OH-]) in the solution.
In this case, we have added 0.100 mol of NaOH, so the concentration of hydroxide ions will be the same. We now need to find the volume of the solution at the equivalence point.
The volume can be calculated by finding the total volume of the acetic acid solution and the volume of NaOH solution required to reach the equivalence point.
Let's assume we need to add V mL of NaOH solution to reach the equivalence point. The total volume of the solution will then be 25.0 mL + V mL.
At the equivalence point, the moles of hydroxide ions provided by NaOH will be equal to the moles of acetate ions produced from the reaction of acetic acid.
moles of NaOH = moles of acetate ions
0.100 mol = (V mL / 1000 L) × 0.100 M
Solving for V:
V mL = (0.100 mol) × (1000 L / 0.100 M) = 1000 mL
So, at the equivalence point, the volume of the solution will be 25.0 mL + 1000 mL = 1025 mL.
Now, let us calculate the concentration of hydroxide ions ([OH-]) in the solution:
[OH-] = (0.100 mol) / (1025 mL / 1000 L) = 0.0976 M
Finally, we can find the pOH:
pOH = -log10(0.0976) ≈ 1.01
Since pH + pOH = 14 (at 25°C), we can calculate the pH at the equivalence point:
pH = 14 - 1.01 ≈ 12.99
Therefore, the pH at the equivalence point of the titration is approximately 12.99.
You must recognize what you have at the equivalence point.
CH3COOH + NaOH ==> CH3COONa + H2O.
At the equivalence point, we have converted all of the CH3COOH to CH3COONa (sodium acetate) so we have M x L = 0.1 x 0.025 = 0.0025 moles sodium acetate. That is in a volume of 50 mL (the other 25 came from 25 mL of 0.1 M NaOH). The pH will be determined by the hydrolysis of water. If we call acetate, Ac^-, and acetic acid HAc, then
Ac^- + HOH ==> HAc + OH^-
Kb for acetate = (Kw/Ka) = (HAc)(OH^-)/(Ac^-).
Set up an ICE chart and solve.
Post your work if you get stuck.
help me solve "use the results of your experiment to determine the various endpoints and then plot a titration curves expected of the two indicators, volume of NaOH (x-axis) pH range of indicators (y-axis)
results are :for bromothymol blue indicator average title volume is27.3 and for phenolphthalein indicator average average titre volume is 27.77