Answer to Question #73728 in General Chemistry for aleia

Question

The distinctive odor of vinegar is due to acetic acid, HC2H3O2. Acetic acid reacts with sodium hydroxide in the following fashion: HC2H3O2(aq) + NaOH(aq) H2O(l) + NaC2H3O2(aq). If 2.79 mL of vinegar requires 34.9 mL of 0.1001 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar? (1 L = 1.0567 qt)

g/qt

Answer 1

CH3COOH (aq) + NaOH (aq)  H2O (l) + CH3COONa
1. Find chemical amount of sodium hydroxide(NaOH):
1000mL contains 0.1001 mole of NaOH;
34.9 mL contains x mole of NaOH;
1000/34.9 = 0.1001/x;
x = 0.0035;
n(NaOH) = 0.0035 mole.

2. The equivalence point in a titration is reached when substances react with each other in equivalent amounts.
Find chemical amount of acetic acid (CH3COOH):
According to equation 1 mole of acetic acid reacts with 1 mole of sodium hydroxide, i.e. n(CH3COOH) = n(NaOH);
n(CH3COOH) = 0.0035 mole.

3. Find mass of acetic acid in 2.79 mL of vinegar:
m = Mn;
M(CH3COOH) = 122+14+162 = 60 (g/mol);
M(CH3COOH) = 600.0035 = 0.21 (g).

4. Find m(CH3COOH) in 1.00 qt of vinegar:
1 L = 1.0567 qt;
2.7910-3 L = y qt;
1/2.7910-3 =1.0567/y;
y= 0.0029 qt.
0.0029 qt of vinegar contains 0.21 g of acetic acid;
1 qt of vinegar contains z g of acetic acid;
0.0029/1 = 0.21/z;
z = 72.41 (g/qt).

Answer: 72.41 g/qt