Forever in debt if you answer this!!!!
Titration of NaOH with Vinegar (acetic acid). Vinegar is in the burette, NaOH in the flask.
I used NaOH with concentration 0.10 mol -1 and i used 25 ml of it.
My Titre of vinegar used to reach equivalence point of NaOH was 21 ml.
The vinegar was DILUTED 1 in 5.
I HAVE TO FIND (A) AND (B)
(a) molarity of diluted vinegar solution (which is 1 in 5 - i did 50 ml vinegar with 200 ml water = 250 ml)
(b) Molarity of original vinegar solution (undiluted 50ml)
MY SOLUTIONS ====== ARE THEY RIGHT METHOD?
(a) C1 X V 1 = C2 X V2
0.10 (concentration of NaOH) X 25 (ml of NaOH) = C2 (? concentration of diluted vinegar) X 21 (titre of vinegar used)
0.10 X 25 / 21 = C2
0.1190 ..... mol -1
(b) since i diluted the vinegar 4 times its volume .....
0.1190 .... X 4 (or is it 5? ) = 0.476 (this is the concentration of undiluted vinegar)
is this right ?
Also, is 0.10 Mol -1 the same as 0.100 mol -1 ???
Titration of NaOH with Vinegar (acetic acid). Vinegar is in the burette, NaOH in the flask.
I used NaOH with concentration 0.10 mol -1 and i used 25 ml of it.
My Titre of vinegar used to reach equivalence point of NaOH was 21 ml.
The vinegar was DILUTED 1 in 5.
I HAVE TO FIND (A) AND (B)
(a) molarity of diluted vinegar solution (which is 1 in 5 - i did 50 ml vinegar with 200 ml water = 250 ml)
(b) Molarity of original vinegar solution (undiluted 50ml)
MY SOLUTIONS ====== ARE THEY RIGHT METHOD?
(a) C1 X V 1 = C2 X V2
0.10 (concentration of NaOH) X 25 (ml of NaOH) = C2 (? concentration of diluted vinegar) X 21 (titre of vinegar used)
0.10 X 25 / 21 = C2
0.1190 ..... mol -1
(b) since i diluted the vinegar 4 times its volume .....
0.1190 .... X 4 (or is it 5? ) = 0.476 (this is the concentration of undiluted vinegar)
is this right ?
Also, is 0.10 Mol -1 the same as 0.100 mol -1 ???