10. For the titration of 20.00 mL 0.1000 M lactic acid with 0.1000 M NaOH, calculate the pH:
a. before any titrant is added
b. 10.00 mL prior to the equivalent point
c. at the equivalent point
d. 10.00 mL after the equivalence point.
Plot the points obtained on a graph of pH versus volume of NaOH added and sketch the curve.
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To calculate the pH at various points during the titration process, we need to consider the dissociation of lactic acid (HC3H5O3) and the reaction with sodium hydroxide (NaOH) to form sodium lactate (NaC3H5O3) and water (H2O).
a. Before any titrant is added:
At this point, the solution only contains lactic acid. Since lactic acid is a weak acid, we can assume that it only partially dissociates. Therefore, we can use the equilibrium expression of lactic acid to calculate the pH. The dissociation reaction is:
HC3H5O3 ⇌ H+ + C3H5O3-
However, since the initial concentration of lactic acid is given as 0.1000 M, we can assume that the concentration of H+ is negligible compared to 0.1000 M. Therefore, the pH would be equal to the negative logarithm of the initial lactic acid concentration: pH = -log(0.1000) = 1.000.
b. 10.00 mL prior to the equivalent point:
At this point, some lactic acid has been neutralized by the addition of NaOH, but the stoichiometric equivalence point has not been reached yet. To calculate the pH, we need to consider the balanced equation for the reaction:
HC3H5O3 + NaOH → NaC3H5O3 + H2O
Using the stoichiometry of the reaction, we can determine that 1 mole of lactic acid reacts with 1 mole of NaOH. Therefore, the concentration of lactic acid remaining after the addition of 10.00 mL of NaOH can be calculated as follows:
[HC3H5O3] = (20.00 mL - 10.00 mL) * 0.1000 M / 20.00 mL = 0.0500 M
Since we have a solution of a weak acid and its conjugate base, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log ([C3H5O3-]/[HC3H5O3])
pKa is the acid dissociation constant for lactic acid, which can be found in a reference table or calculated from equilibrium constants. Assuming a pKa of 3.86 for lactic acid, the equation becomes:
pH = 3.86 + log (0.0500/0.0500) = 3.86
Therefore, the pH at this point is 3.86.
c. At the equivalent point:
The equivalent point is reached when the moles of acid and base in the reaction are equal. At this point, all the lactic acid has reacted with the sodium hydroxide. Therefore, the solution only contains the conjugate base, sodium lactate (NaC3H5O3).
Since sodium lactate is the conjugate base of a weak acid, it can undergo hydrolysis to produce hydroxide ions (OH-). The concentration of OH- ions can be calculated using the ratio of sodium lactate to the initial volume of the solution:
[C3H5O3-] = (10.00 mL * 0.1000 M) / 20.00 mL = 0.0500 M
Since we know the concentration of OH-, we can calculate the pOH:
pOH = -log[OH-] = -log(0.0500) = 1.301
Finally, to find the pH at the equivalent point, we can subtract the pOH from 14 (the negative logarithm of the concentration of H+ ions in pure water at 25°C):
pH = 14 - pOH = 14 - 1.301 = 12.699
Therefore, the pH at the equivalent point is approximately 12.699.
d. 10.00 mL after the equivalence point:
At this point, excess NaOH has been added, and the solution contains an excess of hydroxide ions (OH-). We can calculate the concentration of OH- similar to the previous step and then calculate pOH and pH using the same equations.
Plotting the points obtained on a graph of pH versus the volume of NaOH added will give you a titration curve. With these values, you can sketch the curve on a graph. The pH will start at around 1.000, then gradually increase as NaOH is added until it rapidly increases to approximately 12.699 at the equivalent point. After the equivalent point, the pH will continue to increase slightly as excess NaOH is added.