You have to prepare a pH \(=3.50\) buffer, and you have the following 0.10\(M\) solutions available: \(\mathrm{HCOOH}, \mathrm{CH}_{3} \mathrm{COOH}\) , \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{HCOONa}, \mathrm{CH}_{3} \mathrm{COONa}\) , and \(\mathrm{NaH}_{2} \mathrm{PO}_{4} .\) Which solutions would you use? How many milliliters of each solution would you use to make approximately 1 L of the buffer?

Given solutions are: \(\mathrm{HCOOH}\) (Formic acid), \(\mathrm{CH}_{3}\mathrm{COOH}\) (Acetic acid), \(\mathrm{H}_{3}\mathrm{PO}_{4}\) (Phosphoric acid), \(\mathrm{HCOONa}\) (Sodium formate), \(\mathrm{CH}_{3}\mathrm{COONa}\) (Sodium acetate), \(\mathrm{NaH}_{2}\mathrm{PO}_{4}\) (Sodium dihydrogen phosphate) From these solutions, we can identify the pairs of weak acids and their conjugate bases: \(\mathrm{HCOOH}\) and \(\mathrm{HCOONa}\) (formic acid and sodium formate) \(\mathrm{CH}_{3}\mathrm{COOH}\) and \(\mathrm{CH}_{3}\mathrm{COONa}\) (acetic acid and sodium acetate) \(\mathrm{H}_{3}\mathrm{PO}_{4}\) and \(\mathrm{NaH}_{2}\mathrm{PO}_{4}\) (phosphoric acid and sodium dihydrogen phosphate)

We need to find the pKa of these acids and compare it with the desired pH of the buffer which is 3.50. pKa of formic acid (HCOOH): 3.75 pKa of acetic acid (CH3COOH): 4.76 For phosphoric acid (H3PO4), there are three acidic protons, and the pKa values are 2.15, 7.20, and 12.35. Looking at these values, we can see that the pKa of formic acid is closest to the desired pH. Therefore, we will use formic acid and sodium formate to prepare the buffer.

Using the Henderson-Hasselbalch equation for a buffer system: pH = pKa + log10([Base]/[Acid]) For this problem, pH = 3.50 pKa of formic acid = 3.75 Both the formic acid and sodium formate solutions are 0.10 M. Now, we need to find the ratio of the concentrations: 3.50 = 3.75 + log10([Sodium formate]/[Formic acid])

Subtract pKa from pH: -0.25 = log10([Sodium formate]/[Formic acid]) Now solve for the ratio: 10^(-0.25) = [Sodium formate]/[Formic acid]

We want to prepare approximately 1 L of a buffer solution. Let V1 be the volume of formic acid, and V2 be the volume of sodium formate. Their sum should be close to 1 L: V1 + V2 ≈ 1 L Since we know the ratio of concentrations, let x = [Sodium formate]: [Formic acid] = x / 10^(-0.25) The relationship between concentration and volume is as follows: x * V2 = 0.10 * V1 (for Sodium formate) [x / 10^(-0.25)] * V1 = 0.10 * V2 (for Formic acid) Now we can solve for volumes V1 and V2: V2 = (0.10 * V1) / x V1 + [(0.10 * V1) / x] ≈ 1 V1 ≈ 0.923 * x Similarly, we can find V2: V2 ≈ 0.077 * x For preparing buffer solution of approximately 1 L, you can use any value of x for which V1+V2≈1L, as the concentration and ratio will remain the same. For example, you can keep x=0.1M: V1 ≈ 0.923 L V2 ≈ 0.077 L So, you would need approximately 923 mL of 0.10 M formic acid solution and 77 mL of 0.10 M sodium formate solution to prepare approximately 1 L of pH = 3.50 buffer.