Prepare 500mL of a 0.200 M acetate buffer at pH 4.90 using only pure acetic acid, 3M NaOH, and water.
1) Add acetic acid to ~400mL of water in a 500 mL beaker. How many grams of acetic acid are needed?
2)Add 3 M NaOH solution until pH is 4.90. What volume of NaOH is needed? (ignore activity coefficients)
To answer the first question and determine the amount of acetic acid needed, we need to use the formula for molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
In this case, we want to find the moles of acetic acid required, so we rearrange the equation:
moles of solute = Molarity x volume of solution (in liters)
First, let's convert the given volume of the final solution from mL to liters:
500 mL = 500 / 1000 = 0.5 L
Next, plug in the values into the formula:
0.200 M x 0.5 L = 0.100 moles of acetic acid
To determine the grams of acetic acid needed, we need to use the molecular weight of acetic acid:
Molecular weight (CH3COOH) = 12.01 + 1.01 + 16.00 + 16.00 + 1.01 + 1.01 = 60.05 g/mol
Now, to calculate the grams of acetic acid needed, we use the formula:
grams = moles x molecular weight
grams = 0.100 moles x 60.05 g/mol = 6.005 grams
Therefore, approximately 6.005 grams of acetic acid are needed.
Moving on to the second question, we want to find the volume of 3 M NaOH solution required to adjust the pH to 4.90.
To achieve this, we need to use a pH meter or pH indicator to measure the initial pH of the acetic acid solution. Let's assume the initial pH is higher than 4.90, in which case we need to add the 3 M NaOH solution.
To calculate the volume of NaOH required, we will use the equation:
Molarity x Volume = moles of solute
First, let's convert the desired volume of the final solution from mL to liters:
500 mL = 500 / 1000 = 0.5 L
Next, we determine the moles of acetic acid needed, using the formula from question 1:
0.200 M x 0.5 L = 0.100 moles of acetic acid
Since we need to add NaOH to reach a pH of 4.90, we can assume that acetic acid is the limiting reagent, meaning all of it will react with the NaOH. Therefore, to neutralize the moles of acetic acid, we need an equal number of moles of NaOH.
Hence, 0.100 moles of NaOH are required.
Now, let's calculate the volume of 3 M NaOH solution we need using the formula:
Volume = moles of solute / Molarity
Volume = 0.100 moles / 3 M = 0.033 L or 33 mL
Therefore, approximately 33 mL of the 3 M NaOH solution is needed to reach a pH of 4.90 in the acetate buffer.
1) To prepare a 0.200 M acetate buffer, we need to calculate the amount of acetic acid (CH3COOH) required to achieve the desired concentration.
The molarity formula is given as:
Molarity (M) = moles of solute / liters of solution
First, let's convert the volume of the solution to liters:
400 mL = 400/1000 = 0.4 L
Since we want a 0.200 M solution, we can determine the moles of acetic acid required using the formula:
moles = Molarity × volume (in liters)
moles of acetic acid = 0.200 mol/L × 0.4 L = 0.08 mol
The molar mass of acetic acid (CH3COOH) is:
C = 12.01 g/mol
H = 1.008 g/mol × 4 = 4.032 g/mol
O = 16.00 g/mol + 1.008 g/mol + 1.008 g/mol = 18.016 g/mol
Total molar mass = 12.01 g/mol + 4.032 g/mol + 18.016 g/mol = 30.058 g/mol
Now, we can determine the mass of acetic acid needed:
mass = moles × molar mass
mass = 0.08 mol × 30.058 g/mol = 2.405 g
Therefore, approximately 2.405 grams of acetic acid are needed to achieve a concentration of 0.200 M in the solution.
2) To adjust the pH of the buffer solution to 4.90, we'll add a 3 M NaOH solution. We need to determine the volume of NaOH required.
It is important to note that acetic acid acts as a weak acid, while NaOH is a strong base. The reaction between them will form sodium acetate (CH3COONa) and water. This reaction is given by the equation:
CH3COOH + NaOH → CH3COONa + H2O
Since we are working with a buffer solution, we want to maintain the acetate ion (CH3COO-) concentration in the solution. To do this, we add an excess of acetic acid initially to ensure the presence of both the weak acid and its conjugate base.
Now, let's calculate the volume of 3 M NaOH solution required to adjust the pH to 4.90.
Assuming the volume of water and acetic acid solution is still 400 mL (0.4 L), and we need to bring the pH to 4.90, we'll use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
where:
pH = 4.90
pKa for acetic acid = 4.76 (at 25°C)
[A-] = concentration of acetate ion (0.200 M)
[HA] = concentration of acetic acid (0.200 M)
Rearranging the equation, we have:
[A-] / [HA] = 10^(pH - pKa)
[0.200 M] / [0.200 M] = 10^(4.90 - 4.76) = 10^0.14 = 1.41
This means that the concentration of acetate ion is 1.41 times higher than the acetic acid concentration. Therefore, the moles of acetate ion equals:
moles of acetate ion = 1.41 × 0.08 mol = 0.1128 mol
To determine the volume of 3 M NaOH solution needed, we can use the formula:
moles = Molarity × volume (in liters)
0.1128 mol = 3 mol/L × volume (in liters)
volume (in liters) = 0.1128 mol / 3 mol/L = 0.0376 L
Finally, converting the volume to milliliters:
volume (in mL) = 0.0376 L × 1000 mL/L = 37.6 mL
Therefore, approximately 37.6 mL of 3 M NaOH solution is needed to adjust the pH to 4.90 in the 500 mL buffer solution.
You want b/a ratio to be
4.90 = pKa + log(b/a)
I used 4.74 for pKa and came up with b/a = 0.16 approximately or
base = 0.16*acid
base + acid = 0.2
Solve the two equation simultaneously to obtain acid = ??
base = ??
Then you can answer 1). That will be M/2 = moles and moles/molar mass = grams.
2. Follows from 1. base = 0.16 x acid moles.