Chemist needs a buffer with pH of 4.5.
how many milliliters of pure CH3COOH (density;1.049g/ml) must be added to 500ml of 0.1M NaOH solution to obtain such a buffer?
The Henderson-Hasselbalch equation correctly uses concentrations; however, since the volumes is the same (because it's the same solution), the volumes cancel and one may use mols. I like to use millimoles.
millimols NaOH = mL x M = 500 mL x 0.1 M = 50
Let HAc = CH3COOH. You will need 50 mmols HAc to neutralize the NaOH you have initially. Then HAc must be added to this to form the buffer. You start, therefore, with 50 mmols Ac^- as base. We add acid to this.
pH = pKaHAc + log (base)/(acid)
4.50 = 4.76 + log [50/(50+x)] where x = mmols HAc that must be added. to the 5o mmols Ac^- That works out to be approx 40 mL. Again, you should get a better number than the estimate. To get the buffer you want, you must add 50 mmols HAc to neutralize the NaOH initially PLUS approx 90 mmols to form the buffer or a total of about 140 mmols HAc. Convert that to grams HAc, then use density to convert to mL HAc.
Post your work if you get stuck.
Why did the chemist get a pet parrot?
Because he wanted someone to talk pH with!
To calculate the amount of CH3COOH needed, we first need to find the number of moles of NaOH present in the solution. Given that the volume of the NaOH solution is 500 mL and the concentration is 0.1 M, we can use the formula:
Number of moles = Concentration (M) x Volume (L)
Number of moles of NaOH = 0.1 M x 0.5 L = 0.05 moles
Now, we need to find out how many moles of CH3COOH are needed to form a buffer with a pH of 4.5. We can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Given that the pH is 4.5, we can rearrange the equation to solve for the [A-]/[HA] ratio:
[A-]/[HA] = 10^(pH - pKa)
Since CH3COOH is a weak acid, the pKa value is usually around 4.8-4.9. Let's assume it is 4.8 for this calculation:
[A-]/[HA] = 10^(4.5 - 4.8) = 0.398
To form a buffer, the [A-]/[HA] ratio should be around 1. Therefore, we need the moles of CH3COOH to be equal to the moles of NaOH.
Let's assume the density of CH3COOH is 1.049 g/mL and calculate the required volume of CH3COOH:
Number of moles of CH3COOH = 0.05 moles
Molar mass of CH3COOH = 60.05 g/mol
Mass of CH3COOH = Number of moles x Molar mass = 0.05 moles x 60.05 g/mol = 3.0025 g
Volume of CH3COOH = Mass/Density = 3.0025 g / 1.049 g/mL ≈ 2.86 mL
Therefore, approximately 2.86 mL of pure CH3COOH needs to be added to 500 mL of 0.1 M NaOH solution to obtain a buffer with a pH of 4.5.
To calculate the number of milliliters of pure CH3COOH needed to create a buffer with a pH of 4.5, we first need to determine the acidic and basic components of the buffer and then apply the Henderson-Hasselbalch equation.
The acidic component of the buffer will be CH3COOH (acetic acid), and the basic component will be NaOH (sodium hydroxide). The Henderson-Hasselbalch equation for a buffer is given by:
pH = pKa + log ([A-] / [HA])
Where:
pH = desired pH of the buffer (4.5)
pKa = pKa value of the acid (acetic acid) - This value is 4.76 for acetic acid (CH3COOH).
[A-] = concentration of the conjugate base (CH3COO-) in the buffer
[HA] = concentration of the acid (CH3COOH) in the buffer
Since we want a buffer with a pH of 4.5, the [A-]/[HA] ratio should be 1. To achieve this, we need to add an equal number of moles of acetic acid and acetate ions to the solution.
Now, let's calculate the moles of acetic acid (CH3COOH) required to make the buffer:
Molarity (M) = Moles / Volume (L)
Number of moles of CH3COOH = Molarity x Volume (in liters)
Number of moles of CH3COOH = 0.1 M x 0.5 L (500 mL converted to liters)
Number of moles of CH3COOH = 0.05 moles
We need an equal number of moles of CH3COO- (acetate ions) to achieve a pH of 4.5, so we also need 0.05 moles of CH3COO-.
Next, let's calculate the mass of CH3COOH required using its molar mass (60.05 g/mol):
Mass (g) = Moles x Molar Mass
Mass of CH3COOH required = 0.05 moles x 60.05 g/mol
Mass of CH3COOH required = 3.0025 grams
Finally, let's convert the mass of CH3COOH to volume using its density:
Volume (mL) = Mass / Density
Volume of CH3COOH required = 3.0025 g / 1.049 g/mL
Volume of CH3COOH required ≈ 2.86 mL
Therefore, to obtain a buffer with a pH of 4.5, you need to add approximately 2.86 milliliters of pure CH3COOH (acetic acid) to a 500 mL solution of 0.1M NaOH.
To find out how many milliliters of pure CH3COOH should be added to the existing solution, we need to calculate the amount required to achieve the desired buffer pH of 4.5.
Here's the step-by-step process to solve this problem:
1. Calculate the number of moles of NaOH in the 500 ml solution:
Moles of NaOH = Volume (in liters) × molarity
Moles of NaOH = 0.5 L × 0.1 M = 0.05 moles
2. Determine the number of moles of CH3COOH required to react with the moles of NaOH:
The balanced chemical equation between NaOH and CH3COOH is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we see that 1 mole of CH3COOH reacts with 1 mole of NaOH.
So, the number of moles of CH3COOH required is also 0.05 moles.
3. Calculate the mass of CH3COOH required using its molar mass:
Molar mass of CH3COOH = 12.01 g/mol (carbon) + 1.01 g/mol (hydrogen) + 12.01 g/mol (carbon) + 15.999 g/mol (oxygen) + 1.008 g/mol (hydrogen) = 60.052 g/mol
Mass of CH3COOH required = Moles of CH3COOH × Molar mass of CH3COOH
Mass of CH3COOH required = 0.05 moles × 60.052 g/mol = 3 g
4. Convert the mass of CH3COOH to volume using its density:
Mass = Volume × Density
Volume = Mass / Density
Volume of CH3COOH required = 3 g / 1.049 g/ml = 2.86 ml (approximately)
Therefore, approximately 2.86 milliliters of pure CH3COOH should be added to the 500 ml of 0.1M NaOH solution to obtain a buffer with a pH of 4.5.