50 ml of a .1103M solution of formic acid is titrated with a .2511M NaOH.

a) What is the pH before and NaOH is added?

b) How many mL of NaOH solution must be added to reach the equivalence pt?

c) what is the pH at the equivalence pt of the titration?

d) If the titration is continued for 40mL beyond the equivalence pt, what is the final pH?

a) To find the pH before NaOH is added, we need to determine the initial concentration of H+ ions in the formic acid solution. Formic acid (HCOOH) is a weak acid that partially dissociates in water, so the initial concentration of H+ ions can be calculated using the equilibrium constant (Ka) expression for formic acid.

The equation for the dissociation of formic acid is:

HCOOH ⇌ H+ + HCOO-

The equilibrium constant expression is:

Ka = [H+][HCOO-] / [HCOOH]

Since the formic acid concentration is given as 0.1103 M, and at the start of the titration, no NaOH is added yet, we can assume that all of the formic acid is still present.

Therefore, [HCOOH] = 0.1103 M, [H+] = 0 M, and [HCOO-] = 0 M.

Now we can calculate the initial H+ concentration using the equilibrium constant expression:

0.1103 = (0)(0) / [HCOOH]

Simplifying the equation gives:

[HCOOH] = 0.1103 M

Therefore, the initial concentration of H+ ions is 0.1103 M, and the pH can be calculated using the equation:

pH = -log[H+]

Substituting the value of [H+] gives:

pH = -log(0.1103) ≈ 0.957

So, the pH before NaOH is added is approximately 0.957.

b) The equivalence point of a titration is the point at which the moles of acid and base are stoichiometrically equivalent. To calculate the volume of NaOH solution required to reach the equivalence point, we need to consider the stoichiometry of the reaction between formic acid and NaOH.

The balanced equation for the reaction is:

HCOOH + NaOH → HCOONa + H2O

From this equation, we can see that one mole of formic acid reacts with one mole of NaOH.

To determine the number of moles of NaOH needed, we can use the equation:

moles NaOH = concentration of NaOH × volume of NaOH solution

Since the concentration of NaOH is given as 0.2511 M and the volume of formic acid solution is 50 mL, we can calculate the number of moles of NaOH needed:

moles NaOH = 0.2511 M × (volume of NaOH solution in L)

To reach the equivalence point, we need the moles of NaOH to be equal to the moles of formic acid. Since the moles of formic acid are equal to its concentration (0.1103 M) multiplied by the volume of formic acid (50 mL), we can set up an equation:

moles NaOH = 0.1103 M × (50 mL / 1000 mL/L)

Simplifying the equation gives:

0.2511 M × (volume of NaOH solution in L) = 0.1103 M × (50 mL / 1000 mL/L)

Solving for the volume of NaOH solution gives:

volume of NaOH solution = (0.1103 M × (50 mL / 1000 mL/L)) / 0.2511 M

volume of NaOH solution ≈ 2.20 mL

Therefore, approximately 2.20 mL of NaOH solution must be added to reach the equivalence point.

c) At the equivalence point, the moles of acid and base are stoichiometrically equivalent, resulting in a neutral solution. The reaction between formic acid and NaOH produces sodium formate (HCOONa) and water. Sodium formate is the conjugate base of formic acid, which is a weak acid. In a neutral solution, the conjugate acid and base ions are in equal concentration, resulting in a pH of 7.

Therefore, at the equivalence point of the titration, the pH is 7.

d) If the titration is continued for 40 mL beyond the equivalence point, it means that excess NaOH has been added to the solution. To determine the final pH, we need to consider the hydrolysis of the sodium formate solution.

Sodium formate (HCOONa) can hydrolyze in water, producing formic acid and hydroxide ions (OH-). The equation for the hydrolysis reaction is:

HCOONa + H2O ⇌ HCOOH + NaOH

Since NaOH has been added in excess, the concentration of the hydroxide ions will be higher than the concentration of hydronium ions (H3O+), resulting in a basic solution.

To calculate the final pH, we need to determine the concentration of hydroxide ions (OH-) resulting from the excess NaOH. In this case, we can use the concentration of NaOH (0.2511 M) and the volume of NaOH added beyond the equivalence point (40 mL).

moles OH- = concentration of NaOH × volume of NaOH solution (beyond equivalence point)

moles OH- = 0.2511 M × (40 mL / 1000 mL/L)

Simplifying the equation gives:

moles OH- = 0.2511 M × (0.04 L)

Next, we need to convert the moles of hydroxide ions to concentration by dividing by the total volume of the solution, which is the sum of the original volume of formic acid (50 mL) and the volume of NaOH added beyond the equivalence point (40 mL).

volume of solution = original volume of formic acid + volume of NaOH (beyond equivalence point)

volume of solution = 50 mL + 40 mL

volume of solution = 90 mL = 0.09 L

Finally, we can calculate the concentration of hydroxide ions (OH-) in the final solution:

concentration of OH- = moles OH- / volume of solution

concentration of OH- = (0.2511 M × (0.04 L)) / 0.09 L

Simplifying the equation gives:

concentration of OH- ≈ 0.1113 M

Since the solution is basic due to the presence of excess hydroxide ions, we can calculate the pOH using the equation:

pOH = -log[OH-]

Substituting the concentration of OH- gives:

pOH = -log(0.1113) ≈ 0.952

To find the final pH, we can use the equation:

pH + pOH = 14

Substituting the value of pOH gives:

pH + 0.952 = 14

Solving for pH gives:

pH ≈ 13.048

Therefore, the final pH, after 40 mL beyond the equivalence point, is approximately 13.048.