calculate the volumes of 0.10 M acetic acid and 0.20 M sodium acetate that are needed to prepare
48-mL buffer solution with pH = 3.70. Ka for acetic acid is 1.75 x 10-5
To calculate the volumes of 0.10 M acetic acid and 0.20 M sodium acetate needed to prepare the buffer solution, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH is the desired pH (3.70 in this case)
pKa is the -log(Ka) of the weak acid (acetic acid)
[A-] is the concentration of the conjugate base (sodium acetate)
[HA] is the concentration of the weak acid (acetic acid)
First, let's calculate pKa from the given Ka value:
pKa = -log(1.75 x 10^-5)
pKa = 4.76
Next, let's substitute the values into the Henderson-Hasselbalch equation and solve for the ratio of [A-]/[HA]:
3.70 = 4.76 + log([A-]/[HA])
Rearranging the equation:
log([A-]/[HA]) = 3.70 - 4.76
log([A-]/[HA]) = -1.06
Next, we convert the logarithmic equation into an exponential equation:
[A-]/[HA] = 10^(-1.06)
[A-]/[HA] = 0.092
Since we know that the total volume of the buffer solution is 48 mL, we can set up the following equation:
0.10 M acetic acid volume + 0.20 M sodium acetate volume = 48 mL
Let's assume the volume of acetic acid is x, then the volume of sodium acetate would be (48 - x).
Plugging in the values we have calculated:
0.10 M x + 0.20 M (48 - x) = 48 mL
Simplifying the equation:
0.10x + 0.20(48 - x) = 48
0.10x + 9.6 - 0.20x = 48
-0.10x = 38.4
x = 384 mL
So, the volume of 0.10 M acetic acid needed to prepare the buffer solution is 384 mL, and the volume of 0.20 M sodium acetate needed is (48 - 384) mL, which is -336 mL.
However, we can't have a negative volume, so it means that the concentration of sodium acetate is too high to prepare a buffer with pH 3.70. Hence, you would need to adjust the concentrations or find different concentrations to obtain the desired pH.