Calculate the pH in the titration of 25 mL of 0.10 M acetic acid by sodium hydroxide after the addition to the acid solution of 35 mL of 0.10 M NaOH.
To calculate the pH in the titration of acetic acid with sodium hydroxide, we need to determine the moles of acetic acid and sodium hydroxide involved in the reaction.
First, let's find the moles of acetic acid:
Moles of acetic acid = volume (in L) × concentration
Moles of acetic acid = 0.025 L × 0.10 mol/L
Moles of acetic acid = 0.0025 mol
Since acetic acid is a weak acid, it undergoes a partial dissociation in water. In this case, it dissociates into acetate ions and hydrogen ions:
CH3COOH ⇌ CH3COO- + H+
Next, let's find the moles of sodium hydroxide:
Moles of sodium hydroxide = volume (in L) × concentration
Moles of sodium hydroxide = 0.035 L × 0.10 mol/L
Moles of sodium hydroxide = 0.0035 mol
Since sodium hydroxide is a strong base, it completely dissociates in water. In this case, it produces hydroxide ions:
NaOH → Na+ + OH-
Now, let's determine the limiting reagent. The moles of acetic acid and sodium hydroxide are in a 1:1 ratio according to the balanced equation:
CH3COOH + NaOH → CH3COONa + H2O
Since the moles of acetic acid (0.0025 mol) are less than the moles of sodium hydroxide (0.0035 mol), acetic acid is the limiting reagent.
To determine the concentration of the acetate ions (CH3COO-) after the reaction is complete, we need to subtract the moles of acetic acid used from the initial moles of acetic acid:
Moles of acetate ions = initial moles of acetic acid - moles of acetic acid used
Moles of acetate ions = 0.0025 mol - 0.0025 mol
Moles of acetate ions = 0 mol
Since there are no moles of acetate ions left after the reaction, the resulting solution only contains sodium acetate and water. Sodium acetate is a salt that completely dissociates into sodium ions (Na+) and acetate ions (CH3COO-) in water.
Now, let's calculate the concentration of the hydroxide ions (OH-) in the solution:
Moles of hydroxide ions = initial moles of sodium hydroxide
Moles of hydroxide ions = 0.0035 mol
Since the moles of hydroxide ions are equal to the moles of sodium hydroxide, the concentration of hydroxide ions is:
Concentration of hydroxide ions = moles of hydroxide ions / volume of solution
Concentration of hydroxide ions = 0.0035 mol / 0.060 L
Concentration of hydroxide ions = 0.058 M
To calculate the pOH of the solution, we can use the formula:
pOH = -log[OH-]
Using the concentration of hydroxide ions, we can find the pOH:
pOH = -log(0.058)
pOH ≈ 1.24
Finally, we can calculate the pH of the solution using the relation:
pH = 14 - pOH
pH = 14 - 1.24
pH ≈ 12.76
Therefore, the pH in the titration of 25 mL of 0.10 M acetic acid with 35 mL of 0.10 M sodium hydroxide after the addition is approximately 12.76.
To calculate the pH in the titration of acetic acid by sodium hydroxide, we need to determine the concentration of the resulting solution after the addition of NaOH. First, let's calculate the number of moles of acetic acid initially present:
Moles of acetic acid = volume (L) × concentration (M)
= 0.025 L × 0.10 M
= 0.0025 mol
Next, let's calculate the number of moles of NaOH added:
Moles of NaOH added = volume (L) × concentration (M)
= 0.035 L × 0.10 M
= 0.0035 mol
Since the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH -> CH3COONa + H2O
We can see that the stoichiometry of the reaction is 1:1. Therefore, the number of moles of acetic acid that reacted with NaOH is also 0.0035 mol.
Now, we need to calculate the number of moles of acetic acid remaining after the reaction:
Moles of acetic acid remaining = Moles of acetic acid initially - Moles of acetic acid reacted
= 0.0025 mol - 0.0035 mol
= -0.001 mol
Since the number of moles cannot be negative, it means that all the acetic acid has reacted, and we only have the resulting sodium acetate (CH3COONa).
To find the concentration of the resulting solution, we need to consider the volumes of both solutions used. The total volume after mixing the acetic acid and NaOH is the sum of their volumes:
Total volume = volume of acetic acid + volume of NaOH
= 0.025 L + 0.035 L
= 0.060 L
Now that we know the total volume and the number of moles of sodium acetate, we can calculate its concentration:
Concentration of sodium acetate = Moles of sodium acetate / Total volume
= 0.0035 mol / 0.060 L
= 0.0583 M
However, sodium acetate is the conjugate base of acetic acid and will hydrolyze in water to produce hydroxide ions (OH-). So, we need to consider the ionization of sodium acetate.
The hydrolysis reaction can be written as follows:
CH3COONa + H2O -> CH3COOH + OH-
Since sodium acetate is a strong electrolyte, it completely ionizes in water. Therefore, the concentration of hydroxide ions in the solution will be equal to the concentration of sodium acetate, which is 0.0583 M.
Finally, we can calculate the pOH (negative logarithm of the hydroxide ion concentration):
pOH = -log10(OH- concentration)
= -log10(0.0583)
= 1.232
Since pH + pOH = 14, we can determine the pH of the solution:
pH = 14 - pOH
= 14 - 1.232
= 12.768
Therefore, the pH in the titration of 25 mL of 0.10 M acetic acid by sodium hydroxide after the addition of 35 mL of 0.10 M NaOH is approximately 12.77.
I will let HAc stand for acetic acid.
millimoles HAc initially = mL x M = 25 mL x 0.1 M = 2.5
millimoles NaOH added = 35 x 0.1 = 3.5
..............HAc + NaOH ==> NaAc + H2O
initial.......2.5.........0..............0.............0
add.......................3.5.................................
change....-2.5......-2.5.........+2.5..........+2.5
equilibrium..0..........1.0...........2.5..........2.5
You see that after the addition of the NaOH, all of the HAc is used and you have an excess of 1 mmol NaOH remaining. The pH is determined by the amount of NaOH. That is in a total volume of 25 mL + 35 mL = 60 mL; therefore, (NaOH) = mmols/mL = 1.0/60 = 0.0167 M.
Thus the (NaOH) = 0.0167 M = (OH-) . Convert to (H^+) from
(H^+)(OH^-) = Kw = 1E-14. You know Kw and OH^-, solve for H^+, then convert to pH from pH = -log (H^+)
Post your work if you get stuck.