Find the pH during the titration of a 20.00mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka=1.54 x10^-5), with the addition of 10 mL of 0.1000 M NaOH solution??

To find the pH during the titration of butanoic acid with NaOH, we need to consider the reaction that occurs between these two solutions. The reaction between butanoic acid (HA) and sodium hydroxide (NaOH) results in the formation of sodium butanoate (NaA) and water (H2O). The balanced chemical equation for this reaction is as follows:

CH3CH2CH2COOH (aq) + NaOH (aq) -> CH3CH2CH2COONa (aq) + H2O (l)

To find the pH, we need to determine the concentration of the resulting solutions.

Step 1: Determine the amount of butanoic acid used in the titration.
Given that the initial volume of butanoic acid is 20.00 mL and the concentration is 0.1000 M, we can calculate the number of moles of butanoic acid as follows:

moles of butanoic acid = volume (L) * concentration (M)
moles of butanoic acid = (20.00 mL / 1000 mL per L) * 0.1000 M

Step 2: Determine the amount of NaOH used in the titration.
Given that the volume of NaOH added is 10 mL and the concentration is 0.1000 M, we can calculate the number of moles of NaOH as follows:

moles of NaOH = volume (L) * concentration (M)
moles of NaOH = (10.00 mL / 1000 mL per L) * 0.1000 M

Step 3: Determine the limiting reagent.
To determine the limiting reagent, we compare the moles of butanoic acid and NaOH. The reactant with fewer moles is the limiting reagent, and the other reactant is in excess.

In this case, we compare the moles of butanoic acid and NaOH. If the number of moles of butanoic acid is less than the number of moles of NaOH, then butanoic acid is the limiting reagent. Otherwise, NaOH is the limiting reagent.

Step 4: Determine the moles of the excess reactant.
If butanoic acid is the limiting reagent, the moles of the excess reactant (NaOH) can be determined as follows:

moles of NaOH (excess) = moles of NaOH - moles of butanoic acid

Step 5: Determine the concentration of the sodium butanoate solution.
Since sodium butanoate (NaA) is formed from the complete reaction of butanoic acid and NaOH, we can assume that the volume remains the same. Therefore, the concentration of the sodium butanoate solution is:

concentration of NaA = moles of NaA / volume (L)

Step 6: Calculate the pH of the resulting solution (sodium butanoate solution).
To find the pH of the sodium butanoate solution, we need to consider the dissociation of the salt in water. Sodium butanoate (NaA) is a salt of a weak acid (butanoic acid) and a strong base (NaOH). It will hydrolyze in water to form butanoic acid and sodium hydroxide. Since butanoic acid is a weak acid, the resulting solution will be basic.

The pH of the resulting solution can be calculated using the Henderson-Hasselbalch equation:

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

Where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base (sodium butanoate), and [HA] is the concentration of the weak acid (butanoic acid).

In this case, we need to calculate the concentration of sodium butanoate ([A-]) and butanoic acid ([HA]).

Now, let's proceed with the calculations.

To find the pH during the titration of butanoic acid with NaOH, we need to calculate the concentration of butanoic acid and the concentration of hydroxide ions at the equivalence point.

First, let's find the initial moles of butanoic acid:

moles of butanoic acid = volume (in L) x concentration
moles of butanoic acid = 0.020 L x 0.1000 M
moles of butanoic acid = 0.002 mol

Since butanoic acid is a weak acid, it partially ionizes in water. We need to calculate the concentration of H+ ions using the acid dissociation constant, Ka:

Ka = [H+][A-] / [HA]

Assuming x is the concentration of H+ ions dissociated from butanoic acid, the concentration of undissociated butanoic acid will be (0.002 - x). We can substitute these values into the Ka expression:

1.54 x 10^-5 = x^2 / (0.002 - x)

Since x is assumed to be small compared to 0.002 (due to its weak acid nature), we can approximate (0.002 - x) as 0.002:

1.54 x 10^-5 = x^2 / 0.002

Rearranging the equation and solving for x, we get:

x^2 = (1.54 x 10^-5)(0.002)
x^2 = 3.08 x 10^-8
x ≈ 5.55 x 10^-4 M

Now, let's find the moles of NaOH added:

moles of NaOH = volume (in L) x concentration
moles of NaOH = 0.010 L x 0.1000 M
moles of NaOH = 0.001 mol

Since NaOH is a strong base, it completely dissociates in water to form OH- ions. Therefore, the concentration of OH- ions will be the same as the concentration of NaOH:

concentration of OH- ions = 0.1000 M

At the equivalence point, the moles of OH- ions will react with the moles of H+ ions. Assuming the volume in the titration flask (20.00 mL of butanoic acid) and the volume of added NaOH (10 mL) are both additive, the total volume will be 30.00 mL or 0.030 L.

Now, let's calculate the concentration of OH- ions and H+ ions at the equivalence point:

Concentration of OH- ions = moles of OH- ions / volume
Concentration of OH- ions = 0.001 mol / 0.03 L
Concentration of OH- ions = 0.0333 M

Since H+ ions react with OH- ions in a 1:1 ratio, the concentration of H+ ions at the equivalence point is also 0.0333 M.

To calculate the pH at the equivalence point, we need to use the equation:

pH = -log[H+]

pH = -log(0.0333)
pH ≈ 1.48

Therefore, the pH during the titration of 20.00 mL of 0.1000 M butanoic acid with 10 mL of 0.1000 M NaOH solution is approximately 1.48.

Butanoic acid we'll call HB and you watch the significant figures.

millimols HB = 20 mL x 0.1 M = 2
mmols NaOH added = 10 mL x 0.1M = 1

..........HB + OH^-==> H2O + B^-
initial...2......0......0
add.............1...........
change....-1....-1......1.....1
equil.....1......0.......1....1.

So you have formed 1 mmol base and have left 1 mmols HB (this is a buffer) so use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log(base)/(acid)
base = 1
acid = 1
pKa = -log Ka
Solve for pH. You should get 4.81