i am given equation
CH3COOH + NaOH ---> NaCH3COO + H2O
25ml of acetic acid solution 0.025L
12ml of mid point of titration
1mol of NaOH
find molar concentration
i use a balanced equation
it's a 1:1 ratio
finding NaOH
molarity = mols/litre = mols/L
molarity= 1mol NaOH/0.012 NaOH 83.33mols/l
im a bit confused...i need to find the molar concentration of acetic acid, but i need to get molarity of NaOH first, but i'm not confident in this answer
3 answers
are you in my Peel online class too?
yes..im doing the acetic acid lab..still
The problem states that the mid-point of the titration is 12 mL (I assume that is 12 mL NaOH) so the equivalence point is 2 x 12 = 24 mL (double the mid-point).
Now you know
0.025 L acid x M acid = 0.024 L base x M base. But you need the M or ONE of them to calculate the other. Or at least you need some other information. I'm unfamiliar with the lab you are doing. Go back through it and see if there is any other information given. If so, I suggest you make a new post of it.
Now you know
0.025 L acid x M acid = 0.024 L base x M base. But you need the M or ONE of them to calculate the other. Or at least you need some other information. I'm unfamiliar with the lab you are doing. Go back through it and see if there is any other information given. If so, I suggest you make a new post of it.