1. A solution is prepared such that it is 0.45 M in formic acid and 0.35 M in sodium formate.
a) Where is this mixture located on a titration curve: before the buffer point, at the buffer point, or after the
buffer point?
b) Use the Henderson-Hasselbach to determine the pH of the solution.
c) Ignoring volume effects, determine the pH of the solution when 0.07 M NaOH has been added
LR=Limiting reactant
HA + OH <--> A- + H2O
I 0.45 0.35 - -
LR-0.35 -0.35 +0.35 -
I 0.10 - 0.35 -
C +x +x -x -
E 0.10-x x 0.35-x -
K is Huge the x is removed thus the assumption is hat x<<0.10 and x<<0.35
K = [A-]/[HA][OH-]
1.047 x 10^10 = [0.35M]/[0.10M][x]
2.7 x 10^-9 = [OH-]
Because [A-] is greater the mixture is located after the 1/2 equivalence point.
b. pOH = -log(2.7 x 10^-90
pOH = 8.56
pH = 14- 8.56
pH = 5.44
I am not sure if i did this problem correctly can some one please check and
I don't know how to start part c of the problem?
Are you sure about part a? I think an error is starting with 0.45M formic acid. Wouldn't you need to start with more formic acid to end up with 0.45M?
If you used the HH equation for part b you should have come up with 3.64 and you're a long way from that. I used 3.75 for pKa formic acid.
c.
.........HA + OH^- --> A^- + H2O
I.......0.45..0.......0.35......
add..........0.072..............
C......-0.072.-0.072..+0.072
E.......0.378...0.....0.0.422
pH = 3.75 + log(0.422/0.378) = ?
You right the pH for b = 3.64. and for c =3.78 i messed up on equation. Thank you for for help
a) We can determine the location of the mixture on the titration curve by comparing the concentrations of the acid (HA) and its conjugate base (A-). In this case, the solution is 0.45 M in formic acid (HA) and 0.35 M in sodium formate (A-).
Before the buffer point, the concentration of HA is higher than that of A-, and after the buffer point, the concentration of A- is higher than that of HA.
Since the concentration of A- is higher in this mixture, it is located after the buffer point.
b) To determine the pH of the solution using the Henderson-Hasselbalch equation, we need to know the pKa value of formic acid. The pKa for formic acid is approximately 3.75.
pH = pKa + log([A-]/[HA])
pH = 3.75 + log(0.35/0.45)
pH = 3.75 + log(0.778)
pH = 3.75 + (-0.109)
pH = 3.64
So, the pH of the solution is approximately 3.64.
c) To determine the pH of the solution when 0.07 M NaOH has been added, we need to consider the reaction between NaOH and formic acid (HA).
HA + OH- -> A- + H2O
The reaction ratio is 1:1, which means that for every 1 mole of NaOH added, 1 mole of HA will be consumed and 1 mole of A- will be formed.
Since we have 0.07 M of NaOH, the same concentration of A- will be formed. Therefore, the new concentration of A- will be 0.35 M + 0.07 M = 0.42 M.
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 3.75 + log(0.42/0.45)
pH = 3.75 + log(0.933)
pH = 3.75 + (-0.031)
pH = 3.72
So, the pH of the solution after adding 0.07 M NaOH is approximately 3.72.
To determine if your calculations for part a and b are correct, let's go through them step by step.
a) To determine where the mixture is located on the titration curve, we need to compare the concentrations of the acid (formic acid) and its conjugate base (sodium formate) in the solution. Since the concentration of formic acid (0.45 M) is greater than the concentration of sodium formate (0.35 M), the mixture is located after the buffer point.
b) To determine the pH of the solution using the Henderson-Hasselbalch equation, we need to know the pKa value of formic acid. The pKa value of formic acid is approximately 3.75.
pH = pKa + log([A-]/[HA])
In this case, [A-] is the concentration of sodium formate (0.35 M) and [HA] is the concentration of formic acid (0.45 M).
pH = 3.75 + log(0.35/0.45)
pH ≈ 3.75 - 0.0969
pH ≈ 3.65
So, the pH of the solution using the Henderson-Hasselbalch equation is approximately 3.65.
c) To determine the pH of the solution when 0.07 M NaOH has been added, we need to consider the reaction between formic acid and NaOH. The balanced equation for this reaction is:
HA + OH- → A- + H2O
From the given information, the initial concentration of formic acid is 0.10 M and the initial concentration of sodium formate is 0.35 M. When 0.07 M NaOH is added, it will react with the formic acid.
Using the stoichiometry of the reaction, we can determine the final concentrations of formic acid, sodium formate, and hydroxide ions.
HA: 0.10 M - 0.07 M (0.03 M remains after reaction with NaOH)
A-: 0.35 M + 0.07 M (0.42 M formed from reaction of NaOH with formic acid)
OH-: 0 + 0.07 M (0.07 M NaOH is added)
Now we can use the Henderson-Hasselbalch equation again to determine the pH of the solution.
pH = pKa + log([A-]/[HA])
pH = 3.75 + log(0.42/0.03)
pH ≈ 3.75 + 1.628
pH ≈ 5.38
So, the pH of the solution after 0.07 M NaOH has been added is approximately 5.38.
Please note that all calculations and assumptions should be verified using the appropriate formulas and constants, and the values used in these explanations are approximate.