Answer:
The pH of the solution at the equivalence point is 10.15.
Step-by-step explanation:
a) To calculate the molar concentration of the weak acid in the original sample, we can use the following equation:
Molar concentration of weak acid = Molar concentration of NaOH * Volume of NaOH / Volume of weak acid
Molar concentration of NaOH is given as 0.1000 M and the volume of NaOH required to reach the equivalence point is 29.75 mL. To convert the volume to liters, we need to divide by 1000:
Volume of NaOH = 29.75 mL / 1000 mL/L = 0.02975 L
The volume of the weak acid sample is 25 mL, which is also equal to 0.025 L. We can now calculate the molar concentration of the weak acid:
Molar concentration of weak acid = 0.1000 M * 0.02975 L / 0.025 L
= 0.118 M
Therefore, the molar concentration of the weak acid in the original sample is 0.118 M.
b) At the equivalence point, the number of moles of NaOH added is equal to the number of moles of weak acid initially present in the sample. We can use this fact to calculate the pH at the equivalence point.
The balanced chemical equation for the reaction between lactic acid and NaOH is:
C3H6O3 + NaOH → C3H5O3Na + H2O
From the equation, we see that one mole of lactic acid reacts with one mole of NaOH. The number of moles of NaOH used to reach the equivalence point is:
Moles of NaOH = 0.1000 M * 0.02975 L = 0.002975 moles
This means that the number of moles of lactic acid in the original sample is also 0.002975 moles.
To calculate the pH at the equivalence point, we need to determine the concentration of the conjugate base of lactic acid, which is lactate. We can assume that all of the lactic acid has been converted to lactate at the equivalence point. The concentration of lactate can be calculated using the following equation:
Ka = [H+][C3H5O3-] / [C3H6O3]
Since we are at the equivalence point, [C3H6O3] = [C3H5O3-], and [H+] = [OH-]. Therefore:
Ka = [OH-][C3H5O3-] / [C3H5O3-]
Simplifying:
Ka = [OH-]
[OH-] = Ka = 1.4 x 10^-4 M
Using the equation for the ion product of water, we know that:
[OH-][H+] = 1.0 x 10^-14 M^2
Therefore:
[H+] = 1.0 x 10^-14 M^2 / [OH-]
= 1.0 x 10^-14 M^2 / 1.4 x 10^-4 M
= 7.14 x 10^-11 M
Finally, we can calculate the pH at the equivalence point using the equation:
pH = -log[H+]
= -log(7.14 x 10^-11)
= 10.15
Therefore, the pH of the solution at the equivalence point is 10.15.