What is the pH of a 100 mL solution of 100 mM acetic acid at pH 3.2 following addition of 5 mL of 1 M NaOH? The pKa of acetic acid is 4.70.
The answer is suppose to be 4.75. It will nice to show the steps on how to get this answer. Thank you
But I keep getting 5.70 instead!?
If you had answered my questions this would have been easier. It is NOT possible to have a 0.1 M pure acetic acid solution with a pH of 3.2; therefore, it MUST be a buffer and that probably is your basic problem. Here is how you do it.
pH = pKa + log (base/acid)
3.2 = 4.70 + log b/a. You work it out and
b/a 0.0316 or base = 0.0316*acid
100 mL of the 0.1 M buffer is 10 millimoles. therefore,solve these two equation simultaneously.
acid + base = 10
acid + 0.0315*acid = 10 and
acid = 10/1.0316 = 9.69 and base = 0.306. That is what's in the 0.1 M buffer that you called a 0.1 M acetic acid solution. Now set up an ICE this way but I will call acetic acid HAc(acid) and acetate is Ac^-(base)
You add 5 mL x 1 M NaOH = 5 millimoles NaOH
...............HAc + OH^- ==> Ac^- + H2O
I...............9.69......0..............0.316
add......................5...........................
C...............-5........-5..............+5
E.............4.69........0..............5.316
pH = pKa + log base/acid
pH = 4.70 + log (5.316/4,69).
Solve and pH = 4.75
I'm confused with the numbers. The pH of a 100 mL acetic acid solution that is 0.1 M (100 mM) is 1.4E-3 for (H^+) or pH of 2.85 and not 3.20. Does this mean the solution is not pure acetic acid?
Also did you make a typo on pKa acetic acid. Is that really 4.75 and no 4.70.
Assuming it is a pure acetic acid solution and that pKa is 4.75, which is what I think you have, I would do this. HAc is acetic acid.
millimols HAc = mL x M = 100 x 0.1 = 10
millimols NaOH added = 5 x 1 = 5
........... HAc + OH ==> Ac^- + H2O
I..............10.........0...........0...........0
add......................5..............................
C............-5..........-5............+5...............
E.............5............0..............5
Plug into the Henderson-Hasselbalch equation like this.
pH = pKa + log (base)/(acid)
pH = 4.75 + log (5/5) = 4.75 + 0 = 4.75
Don't hesitate to correct me if I have made the wrong assumptions.
The pka=4.70
And the answer for new ph= 4.75
Thank you for your help! @DrBob222
To find the pH of the solution after the addition of NaOH, we need to consider the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH). This reaction will result in the formation of sodium acetate (CH3COONa) and water (H2O).
The balanced chemical equation for the reaction is as follows:
CH3COOH + NaOH -> CH3COONa + H2O
To determine the pH of the resulting solution, we need to calculate the amount of acetic acid that reacts with the added NaOH.
Step 1: Calculate the initial amount of acetic acid
Given that the volume of the solution is 100 mL and the concentration of acetic acid is 100 mM (mM = millimolar, which means millimoles per liter), we can calculate the initial amount of acetic acid as follows:
Amount (in moles) = concentration x volume (in liters)
Initial amount of acetic acid = 100 mM x (100 mL / 1000 mL/L) = (100 mM / 1000) mole = 0.1 mole
Step 2: Calculate the amount of NaOH added
The volume of NaOH added is 5 mL, and its concentration is 1 M (M = molar, which means moles per liter).
Amount of NaOH added = 1 M x (5 mL / 1000 mL/L) = (1 M / 1000) mole = 0.005 mole
Step 3: Identify the limiting reactant
To determine the limiting reactant, we need to compare the amounts of acetic acid and NaOH:
- Acetic Acid: 0.1 mole
- NaOH: 0.005 mole
Since the amount of NaOH added is much less than the amount of acetic acid, NaOH is the limiting reactant.
Step 4: Calculate the amount of acetic acid that reacts
The balanced chemical equation shows that one mole of acetic acid reacts with one mole of NaOH to form sodium acetate. This means that 0.005 mole of acetic acid will react with 0.005 mole of NaOH.
Step 5: Calculate the remaining acetic acid
To find the amount of acetic acid remaining after the reaction, subtract the amount that reacted:
Remaining acetic acid = Initial amount - Amount reacted
Remaining acetic acid = 0.1 mole - 0.005 mole = 0.095 mole
Step 6: Calculate the concentration of acetic acid in the final volume
The final volume of the solution is 100 mL + 5 mL = 105 mL (converted to liters: 105 mL / 1000 mL/L = 0.105 L).
Concentration of acetic acid = Remaining acetic acid (in moles) / Final volume (in liters)
Concentration of acetic acid = 0.095 mole / 0.105 L = 0.9048 M
Step 7: Calculate the pKa of acetic acid
The pKa of acetic acid is given as 4.70.
Step 8: Calculate the pH of the solution
The Henderson-Hasselbalch equation relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of the acid to its conjugate base (in this case, acetic acid to acetate).
Henderson-Hasselbalch equation: pH = pKa + log10 (conjugate base / acid)
In our case:
pKa = 4.70
Concentration of acetic acid = 0.9048 M
To calculate the concentration of the conjugate base (CH3COO-) from the concentration of acetic acid, we use the following equation:
Concentration of conjugate base = Concentration of acetic acid x 10^(pKa - pH)
Plugging in the values:
0.9048 M x 10^(4.70 - pH)
For the solution to be at pH 3.2, the concentration of the conjugate base and acetic acid should be equal. Thus, we have:
0.9048 M x 10^(4.70 - 3.2) = 0.9048 M
Now, we solve for pH:
10^(4.70 - pH) = 1
Taking the logarithm to the base 10 of both sides to solve for (4.70 - pH):
4.70 - pH = 0
Rearranging the equation to solve for pH:
pH = 4.70
Therefore, the pH of the solution after the addition of 5 mL of 1 M NaOH is 4.70, not 4.75. It seems there was a mistake in the given answer.