ph at equivalence point when 25 ml of a 0.175M solution of acetic acid is titrated with 0.10M of NaOH at its end point
HAc + NaOH ==> NaAc + H2O
mols HAc to start = M x L = ?
moles NaOH must be the same; therefore, the volume of NaOH used is
mLNaOH x MNaOH = mLHAc x MHAc
mL 0.1M NaOH used = 43.75 mL but you need to confirm that.
Thus the salt formed will have a molarity of 0.004375/(43.75mL + 25.00mL) = approximately 0.07 M.
The pH at the equivalence point will be determined by the hydrolysis of the acetate ion of the salt.
............Ac^- + HOH ==> HAc + OH^-
initial...0.07.............0......0
change......-x.............x......x
equil....0.07-x.............x......x
Kb for the Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.07-x)
Solve for x = (OH^-) and convert to pH.
You need to obtain a better answer for the 0.07M, too.
To calculate the pH at the equivalence point when 25 ml of a 0.175M solution of acetic acid is titrated with 0.10M NaOH at its end point, we need to determine the ratio of moles of acetic acid to moles of NaOH at the equivalence point.
First, we need to calculate the number of moles of acetic acid in the 25 ml solution.
Moles of acetic acid = volume (in liters) x concentration
Moles of acetic acid = 0.025 L x 0.175 mol/L
= 0.004375 mol
Since acetic acid and NaOH react in a 1:1 ratio, the moles of NaOH required to reach the equivalence point would also be 0.004375 mol.
Next, we calculate the volume of NaOH required to reach the equivalence point using the given concentration:
Volume of NaOH = moles of NaOH / concentration
Volume of NaOH = 0.004375 mol / 0.10 mol/L
= 0.04375 L
This means that 0.04375 L (or 43.75 ml) of NaOH is required to reach the equivalence point.
At the equivalence point, all the acetic acid has reacted with the NaOH, resulting in the formation of sodium acetate and water.
Sodium acetate is a salt that is fully dissociated in water, so the resulting solution will be a neutral solution.
Therefore, the pH at the equivalence point is approximately 7.
To determine the pH at the equivalence point of an acid-base titration, you need to consider the stoichiometry of the reaction and the dissociation constant of the acid. In this case, since acetic acid (CH3COOH) is a weak acid, you need to account for its dissociation in water.
To find the equivalence point, you need to determine the number of moles of acetic acid in the 25 mL sample. The number of moles (n) can be calculated using the following equation:
n = C * V
Where:
C = concentration of acetic acid (0.175 M)
V = volume of acetic acid (25 mL = 0.025 L)
n = 0.175 M * 0.025 L = 0.004375 moles
Since acetic acid is a weak acid, it does not completely dissociate in water. Instead, it undergoes partial dissociation according to the following equation:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
At the equivalence point, the moles of acetic acid will react with an equal number of moles of hydroxide ions (OH-) from the sodium hydroxide (NaOH) solution, resulting in the formation of an equal number of moles of acetate ions (CH3COO-). This means that the final concentration of CH3COO- will be equal to the initial concentration of acetic acid, which is 0.175 M.
At this point, the solution will be a buffer solution consisting of the acetate ion and the conjugate base of acetic acid. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the dissociation constant of acetic acid (about 4.76)
[A-] = concentration of the acetate ion (0.175 M)
[HA] = concentration of the undissociated acetic acid (0.175 M)
Using the Henderson-Hasselbalch equation, we have:
pH = 4.76 + log(0.175/0.175)
pH = 4.76 + log(1)
pH = 4.76
Therefore, the pH at the equivalence point, when 25 mL of a 0.175 M solution of acetic acid is titrated with 0.10 M NaOH at its end point, is approximately 4.76.