A 12.6 mL sample of vinegar, containing acetic acid, was titrated using 0.542 M NaOH solution. The titration required 24.1 mL of the base. what is the molar concentration of acid in the vinegar?'
both are monoprotic...
Ma*Va=Mb*Vb
Ma=Mb* Vb/va=.542*24.1/12.6=
You must know that 1 mol acetic acid = 1 mol NaOH
mols NaOH - M x L = ?
mols acetic acid = the same
M acid = mols acid/L acid. Substitute and solve for M acid.
This line
mols NaOH - M x L = ?
should read
mols NaOH = M x L = ?
To find the molar concentration of the acid in the vinegar, we can use the concept of stoichiometry and the volume of the NaOH solution required to neutralize the acetic acid.
Here are the steps to calculate the molar concentration of the acid:
1. Determine the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
2. Calculate the number of moles of NaOH used in the titration:
Number of moles = concentration × volume
Number of moles of NaOH = concentration of NaOH × volume of NaOH solution
Number of moles of NaOH = 0.542 M × 24.1 mL
3. Convert the volume of NaOH solution to liters (L):
24.1 mL ÷ 1000 mL/L = 0.0241 L
4. Calculate the number of moles of acetic acid:
According to the balanced equation, the ratio of moles of NaOH to moles of acetic acid is 1:1.
Number of moles of acetic acid = Number of moles of NaOH
5. Calculate the molar concentration of the acetic acid:
Molar concentration = Number of moles ÷ Volume of vinegar sample
Molar concentration of acetic acid = Number of moles of acetic acid ÷ 0.0126 L
Now, we can put the values in:
Molar concentration of acetic acid = (0.542 M × 0.0241 L) ÷ 0.0126 L
By evaluating the above expression, the molar concentration of the acid in the vinegar is obtained.