A 25.0 mL sample of vinegar (dilute acetic actid, CH3 CO2 H ) is titrated and found to react wit

A 25.0 mL sample of vinegar (dilute acetic actid, CH3 CO2 H...

A 25.0 mL sample of vinegar (dilute acetic actid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} )$ is titrated and found to react with 94.7 mL of 0.200 M NaOH. What is the molarity of the acetic acid solution? The reaction is $$\mathrm{NaOH}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$

A 25.0 mL sample of vinegar (dilute acetic actid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} )$ is titrated and found to react with 94.7 mL of 0.200 M NaOH. What is the molarity of the acetic acid
solution? The reaction is
$$\mathrm{NaOH}(a q)+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Na}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$

Key Concepts

Stoichiometry

Stoichiometry involves using the balanced chemical equation to determine the quantitative relationships between reactants and products. In titrations, these mole ratios are essential for converting between the amount of titrant used and the amount of analyte present.

Molarity

Molarity is a concentration unit defined as the number of moles of solute per liter of solution. It is critical in titration calculations as it allows for the determination of the number of moles of a substance from its solution volume, facilitating the overall quantitative analysis.

Acid-Base Titration

Acid-base titration is an analytical technique used to determine the concentration of an acid or base in a solution by slowly adding a titrant of known concentration until the reaction reaches the equivalence point, at which stoichiometrically equivalent amounts of acids and bases have reacted.

Transcript

00:01 In this problem, we're given a lot of information about a base and an acid and asked to calculate the molarity of the acid.

00:12 To this end, we're going to use the equation for molarity, where n is equal to number of moles divided by volume in liters.

00:22 Notice that this equation can be written in three different ways.

00:26 We can solve for molarity.

00:28 You can solve for moles and write this end.

00:32 As m, excuse me, v times m is equal to the number of moles, or we can solve for volume and find the volume is equal to number of moles by molarity.

00:44 These will serve you well as you move through these.

00:47 Now i'm going to go through this in a lot of detail.

00:50 If you're just here for the solution and you want to see the one -step shortcut way of stoichiometry, feel free, but i will take the time to explain each step conceptually.

01:01 So if you want to skip a head to the end, you can go ahead and do that.

01:06 So we have a lot of information about the space over here.

01:13 Notice we have the volume and the molarity.

01:15 So we're going to use that information to find the missing piece.

01:19 That's the number of moles.

01:21 So our first step is going to be to find the number of moles of naoh.

01:28 The reason why we want moles of naah is because in this reaction, we're told by nature of the reaction that one mole of nah reacts with one mole of acetic acid.

01:44 So that i can use the mole to more ratio to find moles of acetic acid.

02:01 So notice that what that does is it gives us moles of this substance and with the volume, we can then find the molarity that we are after.

02:14 So that will be our final step, is to find the molarity of the acid to h3 -co2h.

02:27 So let's start with that first step.

02:32 I need my volume in liters.

02:35 So this is 0 .0947 liters.

02:40 Malarity, it serves as to remember, can be written as 0 .2 -00 mole.

02:45 For every one liter.

02:49 Notice that when i multiply these together, the liters cancel out nicely, and their product is 0 .0 .0894 moles of sodium hydroxide.

03:13 Okay.

03:14 Now recall that we said that for every one mole of sodium hydroxide, we have one mole of the cetic acid.

03:23 So i don't really need to do any math, i just know that because of this one -to -one ratio that i have just as many moles of acetic acid as i did sodium hydroxyl.

03:43 No math involved.

03:44 I just already know the next step.

03:47 And then now what i have, moles of sodium of acetic acid, volume of acetic acid, i can find the molarity by dividing, given the definition of molarity...

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