The species called glacial acetic acid is 98 % acetic acid...

The species called glacial acetic acid is $98 \%$ acetic acid by mass $(d=1.0542 \mathrm{~g} / \mathrm{mL}) .$ What volume of glacial acetic acid must be added to $100.0 \mathrm{~mL}$ of $1.25 \mathrm{M} \mathrm{NaOH}$ to give a buffer with a $\mathrm{pH}$ of $4.20 ?$

Key Concepts

Buffer Solutions

A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists pH changes when small amounts of acid or base are added. In buffer systems, the presence of both the acid and base components allows the solution to neutralize added H+ or OH–, thereby maintaining a relatively stable pH level.

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 its conjugate base and acid. It is given by the formula pH = pKa + log([conjugate base]/[acid]), and is a crucial tool in designing buffers to achieve a specific pH.

Acid-Base Neutralization

Acid-base neutralization involves the reaction of an acid with a base to produce water and a salt. In the context of buffer preparation, careful addition of a weak acid to a strong base (or vice versa) leads to partial neutralization, forming a conjugate pair that can be used to control the pH of the solution.

Molarity and Concentration Calculations

Molarity (M) is defined as the number of moles of solute per liter of solution. Calculations involving molarity require understanding of how to convert between volume and moles, using the relationship between moles, molarity, and volume, which is essential when preparing or adjusting the composition of a solution.

Density and Purity Considerations

Density and purity are important factors when dealing with concentrated reagents. The density allows one to convert between mass and volume, and the percentage purity indicates how much of the substance is actually the active compound. These factors are essential in accurately determining the number of moles available for reaction during solution preparation.

Transcript

00:01 So we're going to react acetic acid with a strong base.

00:05 So let's start by figuring out how many moles of strong base we're going to use.

00:10 So we've got the molarity and the volume.

00:13 We'll put that into liters.

00:15 And we'll see that we added 0 .125 moles of oh -h -minus.

00:24 Our acid is acetic acid.

00:28 The k -a for that is going to be 1 .8 times 10 to negative.

00:33 5.

00:34 Okay, so when that oh minus reacts with that acetic acid, it's going to make 0 .125 moles of acetate.

00:48 We're also told the ph of our solution.

00:52 Our ph is 4 .20.

00:55 So that gives us a hydrogen ion concentration.

00:59 We have 6 .3 times 10 and negative 5 molar.

01:06 Okay, since we're using less hydroxide than we have acid, we're going to end up with a buffer.

01:13 So we're going to take a look at this equation that works well with buffer solutions.

01:18 Concentration of h plus is ka times the moles of the acid, divided by the moles of our base.

01:28 So we've got our hydrogen ion concentration, 6 .3 times 7 minus 5.

01:35 We've got our ka, and then we can set that equal, i'm sorry, you can move.

01:43 Multiply by the moles of acid over the moles of base.

01:49 So we'll go ahead and solve for that ratio of moles of acid over moles of base.

01:57 And that'll come out to be 3 .5...

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