You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.180 M sodium benzoate

How much of each solution should be mixed to prepare this buffer?

Let x = mL benzoate

Then 100-x = mL acid
--------------------
millimols base = x*0.180
mmoles acid = (100-x)*0.1
4.00 = 4.20 + log(0.180x)/[(100-x)*0.1]
Solve for x and 100-x
I get 26 something for base and about 74 for acid. You should do it more accurately.

I need your helping for assigment of account

Ineed helping full for account

To prepare the buffer solution, you need to mix benzoic acid and sodium benzoate in appropriate amounts to achieve the desired pH. Here is the step-by-step calculation:

Step 1: Calculate the ratio of benzoic acid to sodium benzoate:
The Henderson-Hasselbalch equation for a buffer solution is given by:
pH = pKa + log ([A-]/[HA])
Where [A-] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.

pH = 4.00
pKa = 4.20

Rearrange the equation:
log ([A-]/[HA]) = pH - pKa
log ([A-]/[HA]) = 4.00 - 4.20
log ([A-]/[HA]) = -0.20

Take the antilog of both sides to get rid of the logarithm:
[A-]/[HA] = 10^(-0.20)
[A-]/[HA] ≈ 0.63

So, for every 0.63 moles of sodium benzoate, you need 1 mole of benzoic acid in order to achieve a pH of 4.00.

Step 2: Calculate the moles of benzoic acid and sodium benzoate required:
Molarity (M) = moles (n) / volume (V, in liters)
Moles of benzoic acid (nHA) = MHA × VHA
Moles of sodium benzoate (nA-) = MA- × VA-

In this case, we have:
MHA = 0.100 M (concentration of benzoic acid)
MA- = 0.180 M (concentration of sodium benzoate)
VHA + VA- = 100.0 mL = 0.100 L (volume of the buffer solution)

From the ratio calculated in step 1, moles of benzoic acid and sodium benzoate can be related as follows:
nA- = 0.63 × nHA

Since the total volume is 100.0 mL, we can express VA- in terms of VHA:
VHA = 0.100 L - VA-

Now, we can substitute the equations and solve using the given values:
0.180 M × VA- = 0.63 × (0.100 M × (0.100 L - VA-))

Simplify the equation:
0.180 × VA- = 0.063 - 0.063 × VA-
0.180 × VA- + 0.063 × VA- = 0.063
0.243 × VA- = 0.063
VA- = 0.063 / 0.243
VA- ≈ 0.26 L = 260.0 mL (rounded to three significant figures)

Finally, VHA can be determined using VHA = 0.100 L - VA-:
VHA = 0.100 L - 0.260 L
VHA ≈ -0.16 L = -160.0 mL (rounded to three significant figures)

Since we cannot have a negative volume, it means that the concentration of benzoic acid is higher than necessary to achieve the desired pH. In this case, we would use 160.0 mL of the 0.100 M benzoic acid and 260.0 mL of the 0.180 M sodium benzoate to prepare the pH=4.00 buffer solution.

Note: Make sure to double-check your calculation and ensure you have the given values and formulas correct before preparing the buffer solution.