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.