A biochemist needs 750 mL of an acetic acid-sodium ?cetate buffer with pH 4.50. Solid sodium ac

A biochemist needs 750 mL of an acetic acid-sodium ?cetate...

A biochemist needs $750 \mathrm{~mL}$ of an acetic acid-sodium ?cetate buffer with pH 4.50. Solid sodium acetatè $\left(\mathrm{CH}_{3} \mathrm{COONa}\right)$ and glacial acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ are available Glacial acetic acid is $99 \% \mathrm{CH}_{3} \mathrm{COOH}$ by mass and has a density of $1.05 \mathrm{~g} / \mathrm{mL}$. If the buffer is to be $0.15 \mathrm{M}$ in $\mathrm{CH}_{3} \mathrm{COOH}$, how many grams of $\mathrm{CH}_{3} \mathrm{COONa}$ and how many milliliters of glacial acetic acid must be used?

A biochemist needs $750 \mathrm{~mL}$ of an acetic acid-sodium ?cetate buffer with pH 4.50. Solid sodium acetatè $\left(\mathrm{CH}_{3} \mathrm{COONa}\right)$ and glacial acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ are
available Glacial acetic acid is $99 \% \mathrm{CH}_{3} \mathrm{COOH}$ by mass and has a density of $1.05 \mathrm{~g} / \mathrm{mL}$. If the buffer is to be $0.15 \mathrm{M}$ in $\mathrm{CH}_{3} \mathrm{COOH}$, how many grams of $\mathrm{CH}_{3} \mathrm{COONa}$ and how many milliliters of glacial acetic acid must be used?

Key Concepts

Buffer Solutions

A buffer solution is a mixture that resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) and is critical in maintaining pH in biochemical and chemical experiments.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the conjugate base to the acid. This equation is fundamental for calculating the required amounts of acid and conjugate base needed to achieve a desired pH.

Molarity and Mole Calculations

Molarity is the concentration of a solution expressed as moles of solute per liter of solution. Calculating the number of moles of components in a buffer system in a given volume is essential for determining the masses and volumes of the acid and base needed.

Density and Purity Considerations

When dealing with reagents that are not pure or provided in a specific density form, such as glacial acetic acid, it is important to account for both the density and the purity percentage. These factors allow conversion between volume and mass and ensure that the correct amount of active substance is used in the preparation.

Stoichiometry in Buffer Preparation

Stoichiometry involves the calculation of the quantities of reactants and products in chemical reactions. In buffer preparation, it is used to determine the appropriate ratios and total amounts of the weak acid and its conjugate base needed to achieve a desired buffer concentration and pH.

Transcript

00:01 In this problem, we will be discussing the preparation of an acetic acid sodium acetate buffer.

00:07 We are told that we have the need to prepare 750 milliliters of the buffer.

00:18 We need a ph of 4 .50, and we have glacial acetic acid that is 99 % acetic acid by mass, and it has a density of 1 .05 grams per milwaukee.

01:05 We also have solid and i'm assuming pure sodium acetate available.

01:19 So this is what we have.

01:30 If the buffer is specified to be 0 .15 molar in acetic acid, find the milliliters of acidic acid needed and the mass of sodium acetate needed.

02:04 Okay, that's everything we need.

02:08 So for this problem, let's first figure out, probably go to a little bit of a lot, probably go to a lot, another page.

02:14 We're going to take our malarity.

02:19 I'm just going to write acetic acid acid acid acid acid acid a .a.

02:21 That's going to mean acetic acid for right now just to save time.

02:25 To moles.

02:27 They're going to take our moles of acetic acid to grams.

02:31 Then we're going to do our percent and our density.

02:36 And that will give us our milliliters.

02:39 And second, we're going to take our ph to find our, we're going to use that to find our malaria.

02:55 And then we'll use that in our volume to go to moles and then molar mass to go to grams and that's our general plan.

03:07 Okay, let's begin.

03:11 Okay, we were given 0 .15 molarity, so it's moles per liter of, now i'm just going to write h .a for my acid.

03:28 And we were told that we needed to prepare 0 .750 liters, which means we have 0 .1125 moles of our acid...

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