Posted by Aditya on Friday, March 7, 2008 at 12:33pm.
Acetic acid is CH3COOH (HC2H3O2). Its the right end H that is acidic.
CH3COOH + NaOH ==> CH3COONa + HOH
mols NaOH required = M x L = ??
mols acetic acid must be the same since the reaction is 1 mol CH3COOH to 1 mol NaOH.
Molarity CH3OOH = # mols/L = mols/0.10 = xx.
For percent, convert mols CH3COOH from above to grams. Convert 10.0 mL to grams using the density to give the sample mass.
percent CH3COOH = [grams CH3COOH/mass sample]*100 = ??
Post your work if you get stuck.
Ok, so far this is what i've done.
A. to get moles of NaOH, i multiplied 0.5062M x 0.01658L=8.39x10^-3, then i divided this number by .010L, i got .8392M
B. (8.392x10^-3)x(60.052g HC2H3O2)=.5039g, after this part i got stuck.
You're almost home. You have grams and that is correct except I would have carried it out to another place to get 0.50399 which rounds to 0.5040 g. (The molarity I calculated as 0.83928 which rounds to 0.8393 but I didn't round and re-enter numbers in my calculator.
%acetic acid = [grams/mass sample]*100
You have grams from above.
To obtain mass of the sample,
mass = volume x density
10 cc x 1.006 g/cc = ??
Related Questions
chem-acid-base titrations - Acetic acid (HC2H3O2) is an important component of ...
Chem - Please tell me if these are right! 1.How many milliliters of 0.215 M NaOH...
Chemistry - A 10.0 ml sample of vinegar which is an aqueous solution of acetic ...
chemistry - A 10.0-mL sample of vinegar, which is an aqueous solution of acetic ...
Chemistry - Vinegar is a dilute solution of acetic acid. What is the ...
chemistry - 4. Vinegar is a dilute solution of acetic acid. What is the ...
Chemistry - If a 11.0 mL sample of vinegar (aqueous acetic acid ) requires 19.0 ...
Gen. Chemistry lab - Trying to find mass percent of acetic acid in 25 drops of ....
Chemistry - Commercial vinegar was titrated with NaOH solution to determine the ...
Chemistry - A sample of 25.00 mL of vinegar is titrated with a standard 1.02 M ...
For Further Reading
