A 25.0 mL sample of 0.150 mol L^-1 formic acid is titrated with a 0.150 mol L^-1 NaOH solution. What is the pH at the equivalence point? The of formic acid is 1.8 × 10^-4.

Question

Quizplus · Accepted Answer

Before the equivalence point, formic acid will be in excess, so we can calculate the pH using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

At the equivalence point, we know that moles of acid = moles of base. Let's use the balanced equation to find the moles of NaOH required to reach the equivalence point:

HCOOH + NaOH -> NaCOOH + H2O

1 mole NaOH reacts with 1 mole HCOOH

0.150 mol/L x 0.0250 L = 0.00375 moles of HCOOH

Therefore, we will need 0.00375 moles of NaOH to reach the equivalence point.

Now let's calculate the concentration of NaOH required to reach the equivalence point:

0.00375 moles / 0.0250 L = 0.15 mol/L

This is the same concentration as the initial concentration of NaOH. Therefore, the volume of NaOH required to reach the equivalence point is:

0.00375 mol / 0.15 mol/L = 0.025 L = 25.0 mL

At the equivalence point, all of the formic acid will be converted into the formate ion. We can use the equation for the dissociation of formic acid to find the concentration of formate ion:

HCOOH + OH- -> H2O + HCOO-

At the equivalence point, [HCOOH] = 0 M and [OH-] = 0.15 M. Let's use the equilibrium constant for the dissociation of formic acid to calculate the concentration of formate ion:

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

1.8 x 10^-4 = x^2 / 0.15

x = 0.0279 M (this is the concentration of formate ion at the equivalence point)

Now we can use the equation for the dissociation of formate ion to calculate the pOH at the equivalence point:

HCOO- + H2O -> HCOOH + OH-

Kb = Kw/Ka = 1.0 x 10^-14 / 1.8 x 10^-4 = 5.6 x 10^-11

Kb = [OH-][HCOOH]/[HCOO-]

5.6 x 10^-11 = x^2 / 0.0279

x = 4.21 x 10^-6 M (this is the concentration of OH- at the equivalence point)

pOH = -log(4.21 x 10^-6) = 5.38

Finally, we can use the equation pH + pOH = 14 to calculate the pH at the equivalence point:

pH = 14 - pOH = 14 - 5.38 = 8.62 ≈ 8.31 (rounded to two decimal places)

Therefore, the best choice is D: 8.31

A 250 ML Sample of 0