If you add 5.0 mL of 0.50 M NaOH solution to 20.0 mL to Buffer C, what is the change in pH of
the buffer?
(where buffer C is 8.203 g sodium acetate with 100.0 mL of 1.0 M acetic acid)
I have calculated the pH of buffer C to be 4.74.
Now what? =\
pH = pKa + log ([A-]/[HA])
We ned to determine the concentration of the sodium acetate:
Molar mass CH3COONa = 82.0340 g/mol
Therefore you are adding 1mol CH3COOH per litre = 1.00M
pKa CH3COOH = 4.75
pH = 4.75 + log (1/1)
pH = 4.75 as you got - OK.
Question 2: If you add 5 ml of 0.50M NaOH solution to 20ml of the buffer, what is the change in pH?
The answer should be that the pH does not change very much - that is what a buffer solution is all about. I am sure that there is a simple way to make this calculation - but I do not know it. So let us try and do it from first principles:
You have a buffer solution that is 1.00M in CH3COOH and 1.00M in CH3COONa.
If you add some NaOH you react with the CH3COOH , reducing its concentration and increasing the concentration of CH3COONa
Mol CH3COOH in 20 mL of 1.00M = 20/1000*1 = 0.02mol CH3COOH
Mol CH3COONa in 20mL of 1.00M = 0.02mol CH2COONa
Mol NaOH in 5.0mL of 0.50M = 5/1000*0.5 = 0.0025mol NaOH
The NaOH reacts with the CH3COOH to form 0.0025 mol CH3COONa and 0.0025 mol CH3COOH is removed
You end up with
CH3COOH = 0.02mol - 0.0025 mol = 0.0175mol CH3COOH in 25mL total solution:
Molarity of CH3COOH = 0.0175/25*1000 = 0.70M CH3COOH
CH3COONa = 0.02mol + 0.0025mol = 0.0225 mol CH3COONa in 25mL total solution
Molarity of CH3COONa = 0.0225/25*1000 = 0.9M CH3COONa
pH = pKa + log([A-]/[HA])
pH = 4.75 + log( 0.9/0.7)
pH = 4.75 + log 1.286
pH = 4.75 + 0.11
pH = 4.86
couple someone please still explain? and explain how the 4.74 pH was found? thanks
could someone please still explain? and explain how the 4.74 pH was found? thanks
Well, now it's time to spice things up with some NaOH! Adding NaOH to the buffer will have an effect on the pH, so let's see what happens.
First, let's calculate the number of moles of NaOH added. We can use the equation:
moles = concentration x volume
moles = 0.50 M x 0.005 L
moles = 0.0025 mol
Since NaOH is a strong base, it will react with the acetic acid in the buffer. The reaction will go like this:
NaOH + CH3COOH → CH3COONa + H2O
The moles of acetic acid will be equal to the moles of NaOH, since the reaction is 1:1.
Now, let's calculate the new concentration of the acetic acid. We have 0.0025 mol of acetic acid in 0.12 L (20 mL buffer C + 5 mL NaOH). So the new concentration is:
concentration = moles / volume
concentration = 0.0025 mol / 0.12 L
concentration ≈ 0.0208 M
Now that we have the new concentration, we can calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since acetic acid is a weak acid, we need to consider the dissociation in water. The balanced dissociation equation is:
CH3COOH + H2O ⇌ CH3COO- + H3O+
From the equation, we can see that [A-] = [CH3COO-] and [HA] = [CH3COOH]. Substituting these values into the Henderson-Hasselbalch equation, we get:
pH = 4.74 + log(0.0208 / 0.0208)
pH = 4.74 + log(1)
pH = 4.74 + 0
pH = 4.74
Ta-da! The pH of the buffer remains unchanged at 4.74. So, adding that NaOH didn't really affect the pH in this case. Don't worry, though. Your buffer is still hanging in there like a pH superhero.
To determine the change in pH of the buffer after adding 5.0 mL of 0.50 M NaOH solution, you need to consider the reaction that occurs between the acetic acid and sodium hydroxide.
First, we can write the balanced chemical equation for the reaction:
CH3COOH + NaOH -> CH3COONa + H2O
In this reaction, acetic acid (CH3COOH) reacts with sodium hydroxide (NaOH) to form sodium acetate (CH3COONa) and water (H2O).
Since the initial concentrations of the sodium acetate and acetic acid in buffer C are given, we can use the Henderson-Hasselbalch equation to calculate the initial pH of the buffer:
pH = pKa + log ([base] / [acid])
In this case, acetic acid acts as the acid and sodium acetate acts as the base. The pKa of acetic acid is given as 4.74.
Now, to calculate the concentration of the base ([base]), use the equation:
[base] = (moles of sodium acetate / total volume of the buffer)
The moles of sodium acetate can be calculated by dividing its mass by its molar mass:
moles of sodium acetate = mass of sodium acetate / molar mass of sodium acetate
Next, calculate the concentration of the acid ([acid]) using the equation:
[acid] = (moles of acetic acid / total volume of the buffer)
The volume of the buffer is the sum of the volumes of the acetic acid and sodium acetate solution.
With these concentrations, you can now calculate the initial pH of the buffer using the Henderson-Hasselbalch equation.
After that, you need to consider the addition of 5.0 mL of 0.50 M NaOH solution. This solution contains a known amount of moles of sodium hydroxide. You can calculate the moles of NaOH using the equation:
moles of NaOH = volume of NaOH solution (in liters) x concentration of NaOH
Next, consider the stoichiometry of the reaction between NaOH and CH3COOH. For every one mole of NaOH that reacts, one mole of CH3COOH is consumed. This means that the concentration of acetic acid ([acid]) decreases by an amount equal to the moles of NaOH added.
To calculate the new concentration of the acid, subtract the moles of NaOH from the initial moles of acetic acid, and then divide by the total volume of the buffer.
With the new concentrations of the acid and base, you can calculate the new pH of the buffer using the Henderson-Hasselbalch equation.
Finally, calculate the change in pH by subtracting the initial pH from the new pH.
By following these steps, you can determine the change in pH of the buffer after adding 5.0 mL of 0.50 M NaOH solution to 20.0 mL of Buffer C.