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?
What mass of BaSO4 is formed when 18.7 mL of 0.125 M Ba(NO3)2 solution is added to 16.2 mL of 0.152 M H2SO4 solution?
To solve this problem, we will use the concept of stoichiometry and the equation for formic acid and NaOH reaction.
a) To calculate the pH before NaOH is added, we need to find the concentration of H+ ions in the formic acid solution. Formic acid (HCOOH) is a weak acid, which means it partially dissociates in water. The dissociation reaction is as follows:
HCOOH + H2O ⇌ H3O+ + HCOO-
Since the concentration of formic acid is given as 0.1103 M, the concentration of H+ ions is also 0.1103 M (assuming complete dissociation).
To find the pH, we need to take the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+] = -log(0.1103) = 0.958
Therefore, the pH before NaOH is added is approximately 0.958.
b) To determine the volume of NaOH solution required to reach the equivalence point, we can use the stoichiometry of the reaction. The balanced equation for the reaction between formic acid (HCOOH) and NaOH is as follows:
HCOOH + NaOH → HCOONa + H2O
In this reaction, one mole of formic acid reacts with one mole of NaOH. The mole ratio of formic acid to NaOH is 1:1.
Using the equation: (moles = concentration x volume), we can calculate the moles of formic acid:
moles of formic acid = 0.1103 M x 0.050 L = 0.005515 moles
Since the ratio of moles between formic acid and NaOH is 1:1, we will need the same number of moles of NaOH to reach the equivalence point:
moles of NaOH = 0.005515 moles
Now, we will calculate the volume of NaOH solution required using the concentration of NaOH:
volume of NaOH = moles of NaOH / concentration of NaOH
= 0.005515 moles / 0.2511 M
≈ 0.022 mL
Therefore, approximately 0.022 mL of NaOH solution is required to reach the equivalence point.
c) At the equivalence point, all of the formic acid has reacted with NaOH, resulting in the formation of a sodium formate salt (HCOONa). Sodium formate is a salt of a weak acid and a strong base, so it will produce a basic solution.
Since formic acid is a weak acid, it does not completely dissociate. However, at the equivalence point, the concentration of the acid has been neutralized by an equal amount of NaOH, resulting in a salt solution with a pH greater than 7.
Therefore, the pH at the equivalence point in this titration is greater than 7.
d) If the titration is continued for 40 mL beyond the equivalence point, it means that there is now an excess of NaOH in the solution. We will need to calculate the moles of NaOH that remain after the reaction has reached the equivalence point.
moles of NaOH = concentration of NaOH x volume of NaOH
= 0.2511 M x 0.040 L
= 0.010044 moles
Since the stoichiometry of the reaction is 1:1, the moles of formic acid will be the same as the moles of NaOH, which is 0.010044 moles.
Now, we can calculate the remaining volume of NaOH solution:
volume of NaOH = moles of NaOH / concentration of NaOH
= 0.010044 moles / 0.2511 M
≈ 0.040 L
Since the original volume of the solution is 50 mL and we have added an additional 40 mL of NaOH solution, the total volume of the solution is now 90 mL.
To find the final concentration of formic acid and calculate the final pH, we need to consider the excess NaOH as the limiting reagent.
The remaining volume of formic acid solution is 50 mL - 40 mL = 10 mL. The remaining concentration of formic acid will be:
concentration of formic acid = moles of formic acid / volume of formic acid
= 0.010044 moles / 0.010 L
= 1.0044 M
Now, we can calculate the final concentration of H+ ions and find the final pH:
[H+] = concentration of formic acid = 1.0044 M
pH = -log[H+] = -log(1.0044) = -0.001
Therefore, the final pH after 40 mL beyond the equivalence point is approximately -0.001.