General Chemistry Question?!?!?

[farvabull]

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Hey guys, i'm having trouble w/ this question, it's impossible!!! If u could help me out, or atleast explain how to do it i'd really appreciate it. THX

"Vinegar is a dilute aqueous solution of acetic acid. Acetic acid reacts w/ sodium hydroxide according to the following equation:
CH3COOH (acetic acid) + NaOH yields H20 + CH3COONa (vinegar)

If 2.50 mL of vinegar requires 34.9 mL of 0.0960M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1L sample of vinegar?"

 

[gluon999]

Hey guys, i'm having trouble w/ this question, it's impossible!!! If u could help me out, or atleast explain how to do it i'd really appreciate it. THX

"Vinegar is a dilute aqueous solution of acetic acid. Acetic acid reacts w/ sodium hydroxide according to the following equation:
CH3COOH (acetic acid) + NaOH yields H20 + CH3COONa (vinegar)

If 2.50 mL of vinegar requires 34.9 mL of 0.0960M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1L sample of vinegar?"

using M1V1 = M2V2, find the molarity of acetic acid.

then plug in the molarity of acetic acid in the formula M = moles/liter, find the number of moles of acetic acid. then using the molecular weight, find the number of grams of acetic acid.

 

[aamartin81]

Hey guys, i'm having trouble w/ this question, it's impossible!!! If u could help me out, or atleast explain how to do it i'd really appreciate it. THX

"Vinegar is a dilute aqueous solution of acetic acid. Acetic acid reacts w/ sodium hydroxide according to the following equation:
CH3COOH (acetic acid) + NaOH yields H20 + CH3COONa (vinegar)

If 2.50 mL of vinegar requires 34.9 mL of 0.0960M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1L sample of vinegar?"

Though the concept of this question is likely to appear on the MCAT, it would be unlikely to use such odd numbers.

The concept: At equivalence, all the moles of the substance be titrated (acetic acid in vinegar) have been reacted with the titrant (NaOH). In this example, this is a one to one ratio.

Step 1. 34.9ml NaOH x 0.096moles/1000ml = .00335mol OH = .00335 mol acetic acid

Step 2. .00335 mol acetic acid/2.5ml *1000ml/L = 1.34M or moles/L

Step 3. 1.34 moles acetic acid x 60g/mole = 80.4 grams acetic acid

Hope this helps,

Adam

 

[aamartin81]

using M1V1 = M2V2, find the molarity of acetic acid.

then plug in the molarity of acetic acid in the formula M = moles/liter, find the number of moles of acetic acid. then using the molecular weight, find the number of grams of acetic acid.

This is a condensed form of what I typed, much easier to use!

Adam

 

[farvabull]

Thanks a ton, i'm just blanking tonight

I have another one for you...although u can just hint me to what i should do, cuz i feel bad just asking for the answers...thx

"What mass of NaCl is required to precipitate all the silver ions from 20.0 mL of .100 M AgNo3 solution?"

 

[Reimat]

Thanks a ton, i'm just blanking tonight

I have another one for you...although u can just hint me to what i should do, cuz i feel bad just asking for the answers...thx

"What mass of NaCl is required to precipitate all the silver ions from 20.0 mL of .100 M AgNo3 solution?"

You've got to break these down in a way that lets you cancel units- M = moles/L or mmoles/mL.

Moles = g substance / molecular wt. substance

So, by determining the MW of NaCl, you just need to cancel all the units and end up with g in the numerator. This method of canceling units is how I got A's in physics and chem

 

[farvabull]

You've got to break these down in a way that lets you cancel units- M = moles/L or mmoles/mL.

Moles = g substance / molecular wt. substance

So, by determining the MW of NaCl, you just need to cancel all the units and end up with g in the numerator. This method of canceling units is how I got A's in physics and chem

ok, but how does that relate to my 2nd question?

 

[TicAL]

Thanks a ton, i'm just blanking tonight

I have another one for you...although u can just hint me to what i should do, cuz i feel bad just asking for the answers...thx

"What mass of NaCl is required to precipitate all the silver ions from 20.0 mL of .100 M AgNo3 solution?"

NaCl + AgNO3 --> NaNO3 + AgCl

Since there's a 1:1 ratio of reactants, you only need an equal amount of NaCl as the AgNO3 you start with.

mol NaCl = 0.020 L * 0.100 M = 0.002 mol NaCl

g NaCl = 0.002 mol * 28 g/mol = 0.056 g NaCl

 

[Reimat]

ok, but how does that relate to my 2nd question?

My bad- I figured you were just asking about the math

 

[Gentle]

NaCl + AgNO3 --> NaNO3 + AgCl

Since there's a 1:1 ratio of reactants, you only need an equal amount of NaCl as the AgNO3 you start with.

mol NaCl = 0.020 L * 0.100 M = 0.002 mol NaCl

g NaCl = 0.002 mol * 28 g/mol = 0.056 g NaCl

TicAL...great. Does the AL means Aluminium . ok