If you require 30.01 mL of 0.1798 M NaOH solution to titrate 10.0 mL of HC2H3O2 solution, what is the molar concentration of acetic acid in the vinegar?
HC2H3O2 + NaOH ==> NaC2H3O2 + H2O
millimols NaOH = mL x M = 30.01 x 0.1798 = 5.396
millimols HC2H3O2 = the same since the ratio of base to acid is 1mol to 1 mol.
M HC2H3O2 = mmols/mL = 5.396/10.0 = ?
Oh, vinegar, the condiment that adds flavor to both our food and chemistry questions! To find the molar concentration of acetic acid in the vinegar, we need to do a little titration tango with the NaOH solution.
To start, we know that it took 30.01 mL of 0.1798 M NaOH to neutralize 10.0 mL of HC2H3O2. This means that the moles of NaOH used is equal to the moles of acetic acid in the vinegar.
So, let's calculate the moles of NaOH used:
moles of NaOH = volume of NaOH solution (in L) * molarity of NaOH solution
moles of NaOH = (30.01 mL / 1000 mL per L) * 0.1798 M
moles of NaOH = 0.00539998 moles
Since the mole ratio between NaOH and acetic acid is 1:1, the moles of acetic acid in the vinegar is also 0.00539998 moles.
Now, we need to find the molar concentration of acetic acid in the vinegar:
molarity of acetic acid = moles of acetic acid / volume of vinegar (in L)
molarity of acetic acid = 0.00539998 moles / (10.0 mL / 1000 mL per L)
molarity of acetic acid ≈ 0.539998 M
So, the molar concentration of acetic acid in the vinegar is approximately 0.540 M. Cheers to the chemistry of condiments!
To find the molar concentration of acetic acid in the vinegar, we can use the stoichiometry and volume information from the titration.
First, let's write the balanced equation for the reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH):
HC2H3O2 + NaOH -> NaC2H3O2 + H2O
According to the stoichiometry, one mole of acetic acid reacts with one mole of sodium hydroxide. Therefore, the mole ratio is 1:1.
Given that it took 30.01 mL of 0.1798 M NaOH solution to titrate 10.0 mL of HC2H3O2 solution, we can use the following equation to find the moles of NaOH:
moles NaOH = concentration NaOH x volume NaOH
moles NaOH = 0.1798 mol/L x 0.03001 L
moles NaOH = 0.005396 mol
Since the mole ratio between NaOH and HC2H3O2 is 1:1, the moles of acetic acid in the vinegar are also 0.005396 mol.
Now, let's calculate the molar concentration of the acetic acid in the vinegar:
molar concentration of HC2H3O2 = moles of HC2H3O2 / volume of HC2H3O2
molar concentration of HC2H3O2 = 0.005396 mol / 0.0100 L
molar concentration of HC2H3O2 = 0.5396 M
Therefore, the molar concentration of acetic acid in the vinegar is 0.5396 M.
To find the molar concentration of acetic acid in the vinegar, we can use the concept of stoichiometry and the equation for the reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH):
HC2H3O2 + NaOH → NaC2H3O2 + H2O
Here's how we can calculate the molar concentration of acetic acid in the vinegar:
1. Determine the number of moles of NaOH used in the titration.
- Volume of NaOH solution used = 30.01 mL
- Molar concentration of NaOH = 0.1798 M
- Moles of NaOH = volume (in liters) × molar concentration
=> Moles of NaOH = 30.01 mL ÷ 1000 mL/L × 0.1798 mol/L
2. Determine the number of moles of acetic acid (HC2H3O2) present in the reaction.
- According to the balanced chemical equation, the ratio between NaOH and HC2H3O2 is 1:1. Therefore, the moles of HC2H3O2 are equal to the moles of NaOH in this reaction.
3. Calculate the molar concentration of acetic acid in the vinegar.
- Volume of HC2H3O2 solution used = 10.0 mL (converted to liters)
- Molarity of acetic acid (HC2H3O2) = Moles of HC2H3O2 ÷ Volume of HC2H3O2 solution (in liters)
=> Molarity of acetic acid = Moles of NaOH ÷ (10.0 mL ÷ 1000 mL/L)
You can substitute the values into the equation to calculate the molar concentration of acetic acid in the vinegar.