Calculate amount of 1 M acetic acid and 1M sodium acetate necessary to prepare 50 ml of a 0.15 M buffer at ph 4.80. (pka acetic acid =4.74)
Use the HH equation.
pH = 4.80
pKa = 4.74
Solve for the ratio of (base)/(acid). That is equation 1.
The second equation is (base) + (acid) = 0.15
Solve these two equations for (base) and (acid), then convert to mL by using the dilution formula of
mL1 x M1 = mL2 x M2
mL1 x 1M acid = 50 mL x Mfrom problem.
pH = pKA + log salt/acid
4.80 = 4.74 + log salt/acid
4.80-4.74= log salt/acid
0.06 = log salt/acid
antilog0.06 = salt/acid
1.148 = salt/acid
total=1.148+1=2.148
salt=1.148*0.15/2.148 =0.080moles
acid = 1*0.15/2.148 = 0.0698moles
salt=moles*mol.wt=0.080*82.03=
6.56g
acid = moles*mol.wt=0.0698*60.05=
4.19g
1000ml of acid contains=4.19g of acid
1ml contains=4.19/1ooo
50ml contains=4.19/1000*50= 0.2095g per 50ml
1000ml of salt contains = 6.56g of salt
1ml of salt contains=6.56/1000
50ml of salt contains=6.56/1000*50=0.32g per 50ml
To calculate the amount of 1 M acetic acid and 1 M sodium acetate necessary to prepare a 50 ml buffer at pH 4.80, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
In this equation, pKa is the dissociation constant of acetic acid, [A-] is the concentration of the acetate ion, and [HA] is the concentration of acetic acid.
Given that the desired pH is 4.80 and pKa of acetic acid is 4.74, we can plug these values into the equation to solve for the ratio of [A-]/[HA]:
4.80 = 4.74 + log ([A-]/[HA])
0.06 = log ([A-]/[HA])
To solve for the ratio [A-]/[HA], we need to convert the pH back to [H+] concentration using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-4.80)
[H+] = 1.58 x 10^(-5) M
Since acetic acid is a weak acid, at pH 4.80, most of it will be in the undissociated form, while the majority of the sodium acetate will be dissociated into acetate ions.
Now, let's consider the volume of the buffer solution. The total volume is 50 ml, and we want the final buffer concentration to be 0.15 M. Thus, from the Henderson-Hasselbalch equation, we know that the total concentration of the mixture ([A-] + [HA]) should be 0.15 M.
Let's assume x represents the concentration of acetic acid and (0.15 - x) represents the concentration of acetate ions. Since we want the total buffer concentration to be 0.15 M, we can set up the following equation:
x + (0.15 - x) = 0.15
Simplifying the equation, we find:
0.15 - x = 0.15 - 0.15
-x = -0.15
x = 0.15
Now, let's calculate the amount of acetic acid and sodium acetate needed in moles:
Volume of acetic acid = concentration of acetic acid × volume of buffer
Volume of acetic acid = 0.15 M × 0.050 L
Volume of acetic acid = 0.0075 moles
Volume of sodium acetate = concentration of acetate × volume of buffer
Volume of sodium acetate = (0.15 M - 0.0075 M) × 0.050 L
Volume of sodium acetate = 0.1425 moles
Finally, to calculate the amount in grams, we need to know the molar mass of acetic acid (CH3COOH) and sodium acetate (CH3COONa).
The molar mass of acetic acid is approximately 60.05 g/mol, and the molar mass of sodium acetate is approximately 82.03 g/mol.
Amount of acetic acid = Volume of acetic acid × Molar mass of acetic acid
Amount of acetic acid = 0.0075 moles × 60.05 g/mol
Amount of acetic acid = 0.450 grams
Amount of sodium acetate = Volume of sodium acetate × Molar mass of sodium acetate
Amount of sodium acetate = 0.1425 moles × 82.03 g/mol
Amount of sodium acetate = 11.69 grams
Therefore, to prepare a 50 ml buffer at pH 4.80 with a concentration of 0.15 M, you will need approximately 0.450 grams of 1 M acetic acid and 11.69 grams of 1 M sodium acetate.