A 50.0 mL sample of 0.12 M formic acid, HCOOH, a weak monoprotic acid, is titrated with 0.12 M NaOH. Calculate the pH at the following points in the titration. Ka of HCOOH = 1.8 multiplied by 10-4.
What is the pH when
50.0 mL NaOH is added
60.0 mL NaOH is added
70.0 mL NaOH is added
George, how much of this do you know how to do? This is a lot of work for me just to check your answers and I don't want to work these if you already know how. If you have worked these, post your work and I'll be happy to check your answers. I can do that much faster than I can type.
I'm actually not sure how to do any of it.
To do zero mL base.
HCOOH ==> H^+ + HCOO^-
Prepare an ICE chart and substitute into the Ka expression.
Ka = (H^+)(HCOO^-)/(HCOOH)
Solve for (H^+) and convert to pH.
For all of the others. The equation is
HCOOH + NaOH ==> HCOONa + H2O
1. First determine where the equivalence point is; that is, how many mL NaOH must be added to arrive at the equivalence point. For all mL BEFORE the equivalence point to the following:
2. Determine mole HCOOH you have initially. M x L = moles.
3. Add moles NaOH. M x L = moles
4. An ICE chart is the easiest to use but use information from 2 and 3 to determine the amount of salt (HCOONa) formed and the amount of the acid (HCOOH) remaining unreacted. Plug those values into the Ka expression and solve for H^+, then convert to pH.
At the equivalence point, the pH is determined by the hydrolysis of the salt, HCOONa. It's the HCOO^- part that is hydrolyzed.
HCOO^- + HOH ==> HCOOH + OH^-
Set up an ICE chart for this and plug into the following.
Kb = (Kw/Ka) = (HCOOH)(OH^-)/(HCOO^-)
YOu know Kw and Ka (for formic acid), (HCOOH) = (OH^-) = x and (HCOO^-, the salt) you know from the equivalence point work you did. Solve for x = (OH^-) and convert to pH.
For everything past the equivalence point the pH is determine by the excess of NaOH added past the e.p.
Post your work if you get stuck.
To calculate the pH at different points in the titration, we need to consider the chemical reaction that occurs between formic acid (HCOOH) and sodium hydroxide (NaOH). The balanced equation for this reaction is:
HCOOH + NaOH -> HCOONa + H2O
At the start, before any NaOH is added, the solution consists of 0.12 M formic acid. As NaOH is added, it reacts with formic acid in a 1:1 ratio, forming sodium formate (HCOONa) and water (H2O).
Let's go step by step to calculate the pH at each point in the titration:
1. When 50.0 mL of NaOH is added:
First, we need to find the amount of formic acid that has reacted. Since the concentration of formic acid is 0.12 M and the volume of solution is 50.0 mL, the moles of formic acid present would be (0.12 M) x (0.050 L) = 0.006 moles.
Since formic acid and NaOH react in a 1:1 ratio, the moles of NaOH added would also be 0.006 moles.
The total volume of the solution is the original 50.0 mL + 50.0 mL = 100.0 mL = 0.100 L.
Now, we need to consider the reaction of NaOH with formic acid. NaOH is a strong base, so it will completely dissociate into Na+ and OH- ions. The OH- ions will react with the formic acid, resulting in the removal of H+ ions. This reaction is responsible for the increase in pH.
To calculate the new concentration of formic acid, we subtract the moles of NaOH from the moles of formic acid. The new moles of formic acid would be 0.006 moles - 0.006 moles = 0 moles (since all the formic acid is reacted).
Since the total volume is still 0.100 L, the new concentration of formic acid would be 0 moles / 0.100 L = 0 M.
The pH of a 0 M concentration of formic acid is equivalent to the pOH, which can be calculated using the equation: pOH = -log[OH-].
To find the pOH, we need to determine the concentration of OH- ions in the solution. Since the reaction between NaOH and HCOOH is 1:1, the concentration of OH- ions will be equal to the concentration of NaOH added.
Using the concentration: [OH-] = 0.12 M, we can calculate the pOH: pOH = -log(0.12) ≈ 0.92.
Lastly, to calculate the pH, we can use the equation: pH = 14 - pOH. Substitute the pOH value to get: pH = 14 - 0.92 = 13.08.
Therefore, when 50.0 mL of NaOH is added, the pH of the solution is approximately 13.08.
Similarly, you can follow the same steps to calculate the pH when 60.0 mL and 70.0 mL of NaOH are added in the titration. Just remember to adjust the volume of the solution and the moles of NaOH accordingly.