Question

4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted with NaOH. a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many grams of acetic acid are present in the 5 ml of vinegar solution used in the titration? (molar mass of acetic acid = 60 g/mol). b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in vinegar by mass. 5. 50 ml of H2SO4 is neutralized by 25 ml of 0.7 M NaOH. What is the molarity of H2SO4? Bonus Question: NaOH was standardized by titration against a standard acidic solution of potassium hydrogen phthalate (KHP). 0.4798 g of KHP were dissolved in 100 ml of water to prepare the standard KHP solution. a) What is the molar mass of KHP? (you need to look up its formula and calculate its molar mass from the periodic table). b) If 45.22 ml of sodium hydroxide were required to neutralize the KHP solution, what is the molarity of NaOH? Show all your calculations.

          4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted
with NaOH.
a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many
grams of acetic acid are present in the 5 ml of vinegar solution
used in the titration? (molar mass of acetic acid = 60 g/mol).
b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in
vinegar by mass.

5. 50 ml of H2SO4 is neutralized by 25 ml of 0.7 M NaOH. What is the molarity of
H2SO4?

Bonus Question:

NaOH was standardized by titration against a standard acidic solution of potassium
hydrogen phthalate (KHP).
0.4798 g of KHP were dissolved in 100 ml of water to prepare the standard KHP
solution.
a) What is the molar mass of KHP? (you need to look up its formula and
calculate its molar mass from the periodic table).
b) If 45.22 ml of sodium hydroxide were required to neutralize the KHP
solution, what is the molarity of NaOH?
Show all your calculations.

Show more…

4. Vinegar is a solution of acetic acid. Acetic acid has a 1:1 ratio when reacted
with NaOH.
a. Suppose that the molarity of acetic acid in vinegar is 0.5 M, how many
grams of acetic acid are present in the 5 ml of vinegar solution
used in the titration? (molar mass of acetic acid = 60 g/mol).
b. If vinegar has a density of 1.01 g/mol. Calculate the % of acetic acid in
vinegar by mass.

5. 50 ml of H2SO4 is neutralized by 25 ml of 0.7 M NaOH. What is the molarity of
H2SO4?

Bonus Question:

NaOH was standardized by titration against a standard acidic solution of potassium
hydrogen phthalate (KHP).
0.4798 g of KHP were dissolved in 100 ml of water to prepare the standard KHP
solution.
a) What is the molar mass of KHP? (you need to look up its formula and
calculate its molar mass from the periodic table).
b) If 45.22 ml of sodium hydroxide were required to neutralize the KHP
solution, what is the molarity of NaOH?
Show all your calculations.

Added by Melissa J.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Susan Hallstrom

05/26/2023

Step 1

First, we need to find the moles of acetic acid in 5 ml of vinegar solution. We know the molarity of acetic acid is 0.5 M, so we can use the formula: Moles of acetic acid = Molarity × Volume (in liters) Moles of acetic acid = 0.5 M × (5 ml × (1 L / 1000 Show more…

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Transcript

00:01 Okay, let's take a look here.

00:02 So we're asked to consider vinegar, which is acetic acid, and we're told it has a molar mass of 60 grams per mole.

00:19 This is given.

00:22 And it's a one -to -one ratio with sodium hydroxide.

00:31 Okay, question one, number of grams of acetic acid.

00:52 Well, i, okay.

00:55 Number of grams of acetic acid in vinegar.

00:59 And we are given that the molarity is 0 .5 molarity, which is moles per liter.

01:12 And we have 5 milliliters.

01:16 Okay, 5 milliliters times 1 ,000 milliliters per liter times 0 .5 moles per liter times 60 grams per mole.

01:31 Let's do it.

01:33 5 times 0 .5 times 60 divided by 1 ,000 is 0 .15 grams of acetic acid.

01:49 And since we're given so many things with one significant figure, we have to report this to one significant figure equals 0 .2 grams of acetic acid.

02:06 That's the first question.

02:09 For b, if the density equals 1 .01 grams per mole.

02:18 And that's not right.

02:20 It should be 1 .01 grams per milliliter.

02:30 We're asked to find the percent by mass.

02:37 Okay.

02:38 I have 0 .15 grams per 5 milliliters times 1 milliliter is 1 .01, 1 .015 grams in 5 milliliters.

03:10 5 milliliters times 1 .01 grams per milliliter equals 5 .01 grams of solution.

03:44 My percent will be 0 .15 grams divided by 5 .0, that should be 5.

03:54 5 grams equal times 100 equals 2 .97 is 3 .0.

04:23 Again, to one significant figure.

04:27 Our bonus question, we're told we have 0 .4798 grams of khp in 100 mils of water...

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