Question

You need to prepare an acetate buffer of pH 5.47 from a 0.809 M acetic acid solution and a 2.20 M KOH solution. If you have 680 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.47 ? The pa of acetic acid is 4.76. Be sure to use appropriate significant figures.

Answer 1

To create an acetate buffer with a pH of 5.47 using 680 mL of 0.809 M acetic acid and 2.20 M KOH, one must add 1283 mL of the KOH solution.

Step-by-step explanation:

To prepare an acetate buffer with a pH of 5.47 from a 0.809 M acetic acid solution and a 2.20 M KOH solution, we can use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). The pKa of acetic acid is given as 4.76.

Rearranging the equation to solve for the ratio of acetate ion ([A-]) to acetic acid ([HA]) gives us: [A-]/[HA] = 10^(pH - pKa). Substituting the given pH and pKa, we get [A-]/[HA] = 10^(5.47 - 4.76) = 10^0.71 = 5.13. This ratio tells us that for every mole of acetic acid (HA), we need 5.13 moles of acetate ion (A-).

Since we have 680 mL of a 0.809 M solution of acetic acid, the amount of acetic acid present is 0.809 mol/L * 0.68 L = 0.55012 mol. To achieve the desired ratio, we need 0.55012 mol * 5.13 = 2.823 mol of KOH. Given that our KOH solution is 2.20 M, the volume required can be calculated as volume = moles/concentration = 2.823 mol / 2.20 mol/L = 1.283 L or 1283 mL.